Quadratic Formula Calculator: Solve Any Quadratic Equation

Table of Contents - Quadratic Formula

How to Use This Calculator
Understanding the Quadratic Formula
How to Use the Quadratic Formula Manually
Real-World Applications
Common Mistakes and How to Avoid Them
Related Topics
How This Calculator Works
FAQs

How to Use This Calculator - Quadratic Formula

Enter the coefficients from your quadratic equation ax² + bx + c = 0. Put the number in front of x² in the "a" field, the number in front of x in the "b" field, and the constant in the "c" field.

For example, if your equation is 2x² - 5x + 3 = 0, enter a = 2, b = -5, c = 3. Don't forget to include negative signs where they belong.

Click "Solve" to see both solutions (if they exist). The calculator shows you the discriminant, tells you what type of solutions you have, and walks through the formula step-by-step.

Understanding the Quadratic Formula

The quadratic formula is one of the most famous formulas in all of mathematics: x = (-b ± √(b² - 4ac)) / (2a). It solves any quadratic equation ax² + bx + c = 0, no matter how complicated.

Why is this formula so important?

Not every quadratic factors nicely. Some have messy solutions with decimals or even imaginary numbers. The quadratic formula handles all of these cases without breaking a sweat. It's the universal solution method.

Where does it come from?

The formula is derived by completing the square on the general equation ax² + bx + c = 0. Someone did the completing-the-square work once with letters instead of numbers, and we get to use the result forever. That's pretty efficient.

The parts of the formula:

The numerator has two parts: -b (the opposite of the middle coefficient) and ± √(b² - 4ac) (the square root part that creates two solutions). The denominator is always 2a (twice the leading coefficient).

The discriminant:

The expression under the square root, b² - 4ac, is called the discriminant. It tells you what kind of solutions to expect before you even finish calculating. Positive discriminant means two real solutions, zero means one solution, negative means two complex solutions.

Reading the ± symbol:

The plus-minus symbol means you do the calculation twice: once with addition and once with subtraction. This is what gives you two solutions. First solution uses plus, second solution uses minus.

How to Use the Quadratic Formula Manually

Let me walk you through the process with clear examples that cover different scenarios.

Example 1: Two real solutions

Solve: x² - 5x + 6 = 0

Step 1: Identify a, b, and c a = 1 (coefficient of x²) b = -5 (coefficient of x) c = 6 (constant term)

Step 2: Write out the formula x = (-b ± √(b² - 4ac)) / (2a)

Step 3: Substitute the values x = (-(-5) ± √((-5)² - 4(1)(6))) / (2(1)) x = (5 ± √(25 - 24)) / 2 x = (5 ± √1) / 2 x = (5 ± 1) / 2

Step 4: Calculate both solutions First solution (using +): x = (5 + 1) / 2 = 6 / 2 = 3 Second solution (using -): x = (5 - 1) / 2 = 4 / 2 = 2

Solutions: x = 3 or x = 2

Step 5: Check your work For x = 3: (3)² - 5(3) + 6 = 9 - 15 + 6 = 0 ✓ For x = 2: (2)² - 5(2) + 6 = 4 - 10 + 6 = 0 ✓

Example 2: One repeated solution

Solve: x² + 6x + 9 = 0

Step 1: Identify coefficients a = 1, b = 6, c = 9

Step 2: Apply formula x = (-6 ± √(6² - 4(1)(9))) / (2(1)) x = (-6 ± √(36 - 36)) / 2 x = (-6 ± √0) / 2 x = (-6 ± 0) / 2

Step 3: Calculate x = -6 / 2 = -3

This gives one solution: x = -3 (repeated)

The discriminant was zero, which always means one repeated solution.

Example 3: Irrational solutions

Solve: x² - 4x + 1 = 0

Step 1: Identify coefficients a = 1, b = -4, c = 1

Step 2: Apply formula x = (4 ± √(16 - 4)) / 2 x = (4 ± √12) / 2 x = (4 ± 2√3) / 2 x = 2 ± √3

Step 3: Write both solutions x = 2 + √3 ≈ 3.732 x = 2 - √3 ≈ 0.268

These are exact irrational solutions.

Example 4: Leading coefficient not 1

Solve: 2x² + 7x - 4 = 0

Step 1: Identify coefficients a = 2, b = 7, c = -4

Step 2: Apply formula carefully x = (-7 ± √(7² - 4(2)(-4))) / (2(2)) x = (-7 ± √(49 + 32)) / 4 x = (-7 ± √81) / 4 x = (-7 ± 9) / 4

Step 3: Calculate both solutions x = (-7 + 9) / 4 = 2 / 4 = 1/2 x = (-7 - 9) / 4 = -16 / 4 = -4

Solutions: x = 1/2 or x = -4

Example 5: Complex solutions

Solve: x² + 2x + 5 = 0

Step 1: Identify coefficients a = 1, b = 2, c = 5

Step 2: Apply formula x = (-2 ± √(4 - 20)) / 2 x = (-2 ± √(-16)) / 2 x = (-2 ± 4i) / 2 x = -1 ± 2i

Step 3: Write solutions x = -1 + 2i or x = -1 - 2i

These are complex conjugate pairs.

Example 6: Negative leading coefficient

Solve: -x² + 3x + 10 = 0

Step 1: Identify coefficients (don't forget the negative!) a = -1, b = 3, c = 10

Step 2: Apply formula x = (-3 ± √(9 - 4(-1)(10))) / (2(-1)) x = (-3 ± √(9 + 40)) / (-2) x = (-3 ± √49) / (-2) x = (-3 ± 7) / (-2)

Step 3: Calculate both solutions x = (-3 + 7) / (-2) = 4 / (-2) = -2 x = (-3 - 7) / (-2) = -10 / (-2) = 5

Solutions: x = -2 or x = 5

Real-World Applications

Physics projectile problems:

When you throw something upward, its height follows h = -16t² + v₀t + h₀. The quadratic formula tells you when the object hits the ground (h = 0) or reaches a certain height.

Business break-even analysis:

Profit functions are often quadratic: P = -x² + 50x - 200. The quadratic formula finds the production levels where profit equals zero (break-even points).

Engineering trajectory calculations:

Bridges, arches, and parabolic paths all involve quadratic equations. Engineers use the formula to find key points like where supports should be placed.

Geometry area problems:

If you need to find dimensions where area equals a specific value, you often end up with quadratic equations. The formula solves for the unknown dimension.

Optics and lens formulas:

Focal length calculations in physics involve quadratic equations when dealing with multiple lenses or curved mirrors.

Economics supply and demand:

Market equilibrium points where supply equals demand sometimes involve solving quadratic equations when the relationships aren't linear.

Computer graphics:

Calculating where a ray intersects a parabolic surface uses the quadratic formula. Video game engines and 3D rendering software use this constantly.

Common Mistakes and How to Avoid Them

Mistake 1: Forgetting the negative sign in front of b

Wrong: Using b instead of -b in the numerator

Right: The formula starts with -b, not b. If b is already negative, -b becomes positive. For b = -5, you write -(-5) = +5.

Why it happens: The formula has a minus sign built in. We forget it's there and just plug in the number as is.

Mistake 2: Missing the 4ac under the square root

Wrong: Writing √(b² - ac) instead of √(b² - 4ac)

Right: The discriminant is b² - 4ac. That 4 is crucial and comes from completing the square derivation.

Why it happens: Writing too fast or misremembering the formula. Double-check before calculating.

Mistake 3: Forgetting to divide by 2a

Wrong: Calculating -b ± √(b² - 4ac) and stopping there

Right: The entire expression must be divided by 2a. Both parts of the numerator get divided, not just one.

Why it happens: The denominator is easy to overlook, especially when it's just 2. Always write the full formula before substituting.

Mistake 4: Distributing the division incorrectly

Wrong: For (-6 ± 4) / 2, writing -6/2 ± 4 instead of (-6 ± 4) / 2

Right: The entire numerator (-6 ± 4) is divided by 2. Calculate the ± operations first, then divide.

Why it happens: Misunderstanding order of operations. Parentheses in the numerator mean everything inside divides by the denominator.

Mistake 5: Sign errors when substituting negative coefficients

Wrong: For b = -3, writing -b = -3 instead of -b = 3

Right: If b is negative, then -b is positive. Multiply the signs carefully.

Why it happens: Rushing through sign changes. Write it out: -b = -(-3) = +3.

Mistake 6: Arithmetic errors with the discriminant

Wrong: Calculating (-5)² as -25 instead of 25

Right: Squaring a negative number always gives a positive result. (-5)² = (-5)(-5) = +25.

Why it happens: Confusing squaring with just putting a negative sign on the number.

Mistake 7: Not simplifying radicals

Wrong: Leaving the answer as (4 ± √12) / 2

Right: Simplify √12 = 2√3, giving (4 ± 2√3) / 2 = 2 ± √3

Why it happens: Stopping too soon. Always simplify radicals and reduce fractions when possible.

How This Calculator Works

Step 1: Parse and validate input

Read a, b, c values
Check that a ≠ 0 (must be quadratic)
Validate all inputs are numbers

Step 2: Calculate discriminant

discriminant = b² - 4ac
This determines solution type

Step 3: Determine solution type

if discriminant > 0: two real solutions
if discriminant = 0: one repeated solution
if discriminant < 0: two complex solutions

Step 4: Calculate solutions

sqrt_discriminant = √|discriminant|
if discriminant >= 0:
    x₁ = (-b + sqrt_discriminant) / (2a)
    x₂ = (-b - sqrt_discriminant) / (2a)
else:
    real_part = -b / (2a)
    imaginary_part = sqrt_discriminant / (2a)
    x₁ = real_part + imaginary_part×i
    x₂ = real_part - imaginary_part×i

Step 5: Simplify and format

Simplify radicals if possible
Reduce fractions
Format complex numbers as a + bi

Step 6: Display results

Show discriminant value
Show solution type
Show both solutions with steps
Provide decimal approximations

FAQs

What is the quadratic formula?

It's x = (-b ± √(b² - 4ac)) / (2a), a formula that solves any quadratic equation ax² + bx + c = 0. It works for every quadratic, no exceptions.

When should I use the quadratic formula?

Use it when factoring is difficult or impossible, when you need exact decimal answers, or when you want a reliable method that always works regardless of the coefficients.

What does the ± symbol mean?

It means "plus or minus." You calculate the formula twice: once adding the square root part, once subtracting it. This gives you two solutions.

What if my discriminant is negative?

You'll get complex (imaginary) solutions involving i. The formula still works; you just need to work with √(negative number) = i√(positive number).

Can the quadratic formula give me one answer?

Yes, when the discriminant equals zero, both the + and - versions give the same answer. This is called a repeated or double root.

Do I always get two solutions?

Yes, but they might be: two different real numbers, one repeated real number, or two complex conjugates. Every quadratic has exactly two solutions (counting multiplicities).

What if a = 0?

Then it's not a quadratic equation anymore, it's linear. The quadratic formula doesn't apply. Just solve bx + c = 0 normally.

How do I remember the formula?

Many people use a song or rhyme: "x equals negative b, plus or minus the square root, of b squared minus 4ac, all over 2a." Singing it helps!

What's the discriminant again?

It's b² - 4ac, the expression under the square root. It discriminates (tells apart) what type of solutions you'll get: positive for two real, zero for one repeated, negative for complex.

Can I use this formula for equations not equal to zero?

First move everything to one side to get the standard form ax² + bx + c = 0, then use the formula with those coefficients.

What if my equation has fractions?

The formula handles fraction coefficients fine. Just substitute them in. Or multiply the entire equation by a common denominator first to clear fractions.

Why is there a 2a in the denominator?

It comes from completing the square on the general quadratic equation. The factor of 2 appears when you factor out the leading coefficient.

Can this solve cubic or higher degree equations?

No, only quadratics (degree 2). Cubic equations have their own formulas, but they're much more complicated.

What if I get a really messy radical?

Simplify it as much as possible by factoring out perfect squares. √12 = √(4×3) = 2√3. Your final answer will be cleaner.

Do I need to simplify my answer?

Yes, always simplify radicals and reduce fractions. (6 ± 2√3) / 4 should be simplified to (3 ± √3) / 2.

What if both a and b are zero?

Then you just have c = 0, which is only true if c = 0. It's not really an equation to solve.

Can I check my answers?

Absolutely! Substitute each solution back into the original equation. If it makes the equation true (gives 0), your answer is correct.

What's the relationship between solutions and factors?

If x = 2 and x = 5 are solutions, then (x - 2) and (x - 5) are factors. The equation factors as a(x - 2)(x - 5) = 0.

Why learn this if I can just use a calculator?

Understanding the formula helps you understand quadratics deeply, recognize patterns, and solve problems where calculators aren't available or appropriate.

Can the formula fail or give wrong answers?

If you use it correctly with arithmetic done right, it never fails. All errors come from arithmetic mistakes, not the formula itself.

Quadratic Formula Calculator