The quadratic formula is the most reliable tool ever created for solving second-degree equations. When factoring fails — or when guessing integer factors would take forever — this formula delivers exact solutions every single time. Learn its four-step routine, understand the discriminant shortcut, and you will never be stuck on a quadratic again.
What Is the Quadratic Formula?
The formula solves any equation written in standard form:
ax² + bx + c = 0
- a, b, and c are known numbers (with a ≠ 0).
- x is the unknown you are solving for.
The complete formula is:
x = (-b ± √(b^2 – 4ac))/(2a)
The ± symbol means you get two solutions: one from adding the radical term, one from subtracting it.
Before reaching for the formula, it is worth checking whether the equation factors simply — our guide on how to factor polynomials with examples covers that method. For tricky coefficients, the CalcSolver math engine verifies each coefficient before you start calculating by hand.
The Discriminant: Know Your Answers Before You Compute
The expression inside the square root, b^2 – 4ac, is called the discriminant (often written Δ). It predicts the type of solution:
- Discriminant > 0: two distinct real solutions (the parabola crosses the x-axis twice).
- Discriminant = 0: exactly one real solution, a repeated root (the parabola just touches the x-axis).
- Discriminant < 0: no real solutions — the roots are complex/imaginary.
Checking the discriminant first takes five seconds and can save you from a full page of pointless arithmetic.
The 4-Step Solving Framework
- Arrange in standard form: move everything to one side so the equation equals zero.
- Extract coefficients: identify a, b, and c, keeping their signs.
- Substitute: place each value in parentheses inside the formula to prevent sign errors.
- Compute and simplify: evaluate the discriminant, take the square root, and work out both branches.
Example 1: x² − 5x + 6 = 0
Step 1: Identify the coefficients
- a = 1
- b = −5
- c = 6
Step 2: Plug into the formula
x = (−(−5) ± √((−5)² − 4(1)(6))) / (2(1))
Step 3: Simplify step by step
- −(−5) = 5
- (−5)² = 25
- 4 × 1 × 6 = 24
- 25 − 24 = 1
x = (5 ± √1)/2 = (5 ± 1)/2
Step 4: Split into both branches
- Addition branch: x = (5 + 1)/2 = 6/2 = 3
- Subtraction branch: x = (5 − 1)/2 = 4/2 = 2
Solutions: x = 3 and x = 2.
Example 2: 2x² + 3x − 2 = 0 (Fractions and Tricky Signs)
Step 1: Identify the values
- a = 2
- b = 3
- c = −2
Step 2: Substitute into the formula
x = (−(3) ± √((3)² − 4(2)(−2))) / (2(2))
Step 3: Evaluate inside the square root
- 9 − (−16) = 9 + 16 = 25
- √25 = 5
x = (−3 ± 5)/4
Step 4: Evaluate both solutions
- Positive branch: x = (−3 + 5)/4 = 2/4 = 1/2 (0.5 as a decimal)
- Negative branch: x = (−3 − 5)/4 = −8/4 = −2
Solutions: x = 1/2 and x = −2.
Fractional answers come up constantly with this formula. If fraction arithmetic slows you down, review our tutorials on 1/4 as a percent and decimal and what is equivalent to 3/4.
Example 3: Repeated Root — x² + 4x + 4 = 0
Sometimes the discriminant is zero. Compute it first: b² − 4ac = 16 − 16 = 0. One solution only:
x = (−4 ± √0)/2 = −4/2 = −2
Solution: x = −2 (a repeated root). Indeed, x² + 4x + 4 = (x + 2)².
The discriminant is your five-second preview. Compute b^2 – 4ac before doing anything else — it tells you how many solutions to expect and whether the square root will be messy.
Essential Pitfalls to Avoid
- Squaring negatives wrong: write (−5)², not −5². A squared real number is always positive; −5² means −(25).
- Partial division: the denominator 2a divides the entire numerator (−b ± √(…)), not just the radical.
- Missing negative signs: if c = −6, then −4ac becomes −4(a)(−6), a positive addition inside the root. Track every sign.
- Forgetting standard form: an equation like 2x² = 5x − 3 must be rearranged to 2x² − 5x + 3 = 0 before you read off a, b, c.
Always run both branches of the ± and plug each answer back into the original equation. Two quick checks catch almost every sign slip.