Quadratic Equations: Four Ways to Solve, One Formula to Rule Them
A quadratic is any equation with an x-squared term, and its graph is the familiar parabola. There are four ways to solve one — and the quadratic formula always works, even when nothing factors cleanly.
The Short Version
- A quadratic in standard form is ax² + bx + c = 0; its graph is a parabola.
- Solve by factoring when it's clean, by square roots when there's no x-term, and by the quadratic formula always.
- The quadratic formula: x = (−b ± √(b² − 4ac)) / 2a.
- The discriminant b² − 4ac tells you how many real solutions exist — a recurring SAT and ACT question.
A quadratic equation is any equation where the highest power of the variable is two. Graph one and you get a parabola — a smooth U-shaped curve. The solutions to the equation are exactly where that curve crosses the x-axis. The tests ask you to find those solutions, and you have a small toolkit of methods, one of which (the quadratic formula) never fails.
This guide works through all four methods, shows how the discriminant counts solutions without solving, maps the topic across each test, and ends with worked and practice problems calibrated to real test difficulty at Northside Tutoring.
Why Quadratics Matter
Quadratics are the most-tested algebra topic above the linear level. They appear as projectile-motion word problems, area problems, parabola graphs, and pure "solve for x" questions. The SAT in particular leans on quadratics heavily, including questions that test whether you understand the connection between an equation, its factors, and its graph.
Standard Form & the Parabola
Every quadratic can be arranged into standard form: ax² + bx + c = 0, with a ≠ 0. The graph is a parabola that opens up when a > 0 and down when a < 0. The solutions (also called roots or zeros) are the x-values where y = 0 — the x-intercepts.
The two solutions of a quadratic are where its parabola crosses the x-axis.
Method 1: Factoring
If the quadratic factors into two binomials, set each to zero. To solve x² + 5x + 6 = 0, factor into (x + 2)(x + 3) = 0, which means x = −2 or x = −3. This is the fastest method — when it works. (Our factoring guide covers the full technique.)
Method 2: The Square Root Method
When there's no middle (bx) term, isolate the square and take the root of both sides — remembering the ±:
This also handles forms like (x − 3)² = 16, giving x − 3 = ±4, so x = 7 or x = −1.
Method 3: The Quadratic Formula
When nothing factors cleanly, the quadratic formula solves any quadratic in standard form:
Identify a, b, and c from standard form, substitute carefully (watch the signs), and simplify. It's slower than factoring but it never fails, which makes it the safety net on every test.
Completing the square
A fourth method, completing the square, rewrites the quadratic as a perfect square plus a constant. It's how the quadratic formula is derived and how you convert to vertex form — useful, but on test day the formula is usually faster.
The Discriminant: Counting Solutions
The piece under the radical, b² − 4ac, is the discriminant. You don't have to finish solving to know how many real solutions there are:
| Discriminant b² − 4ac | Number of real solutions |
|---|---|
| Positive | 2 real solutions |
| Zero | 1 real solution (a repeated root) |
| Negative | 0 real solutions (2 complex) |
Where You'll See This — Test by Test
No reference sheet provides the quadratic formula — you must memorize it. Quadratics are among the most frequent topics on the SAT and ACT, and the discriminant shows up as its own conceptual question.
Digital SAT
A top-tested topic: factoring, the formula, vertex form, and discriminant questions all appear. Often linked to parabola graphs.
Explore SAT Tutoring → College AdmissionsACT
Frequent. Expect to factor, apply the formula, and occasionally count solutions with the discriminant.
Explore ACT Tutoring → Independent School AdmissionsSSAT
Upper Level introduces solving simple quadratics by factoring and square roots.
Explore SSAT Tutoring → K-12 CurriculumSchool Algebra
The centerpiece of Algebra I and II; foundational for functions, parabolas, and pre-calculus.
Explore Algebra Tutoring →Watch the Lesson
Sometimes a diagram needs a voice. In the short video below, one of our Northside tutors walks through the core idea and works through test-style problems in real time.
Quadratics — In Plain English
A live walkthrough from our tutoring team.
— Featuring a Northside Tutoring instructor
For the developer / editor
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Worked Example Problems
These problems are calibrated to the difficulty you'll actually see on test day. Try each one before opening the solution.
Solve: x² − 7x + 12 = 0.
Show solution
Factor into two numbers that multiply to 12 and add to −7: −3 and −4.
(x − 3)(x − 4) = 0 → x = 3 or x = 4.
Solve: 2x² = 50.
Show solution
Divide by 2: x² = 25. Take the square root of both sides: x = ±5.
Solve x² + 4x − 1 = 0 using the quadratic formula.
Show solution
a = 1, b = 4, c = −1. Discriminant = 16 − 4(1)(−1) = 20.
x = (−4 ± √20) / 2 = (−4 ± 2√5) / 2 = −2 ± √5.
How many real solutions does x² + 3x + 5 = 0 have?
Show solution
Discriminant = b² − 4ac = 9 − 4(1)(5) = 9 − 20 = −11.
Negative discriminant → no real solutions.
A ball's height is h = −16t² + 32t. At what positive time t does it hit the ground (h = 0)?
Show solution
Set 0 = −16t² + 32t = −16t(t − 2).
So t = 0 (launch) or t = 2. The positive landing time is t = 2.
Common Mistakes to Avoid
Three traps that catch students every year
- Dropping the ± when taking a square root. x² = 49 has two solutions, +7 and −7. Forgetting the negative loses half the answer.
- Sign errors in the formula. The formula starts with −b. If b is negative, −b is positive — substitute with parentheses to stay safe.
- Forgetting to set the equation to zero first. Factoring and the formula only work when one side is 0. Rearrange before you start.
Practice Problems — You Try
Three problems below. Work each before checking the solution.
Solve: x² − 9 = 0.
Show solution
x² = 9 → x = ±3.
Solve: x² + 2x − 15 = 0.
Show solution
Factor: two numbers multiplying to −15, adding to 2: 5 and −3.
(x + 5)(x − 3) = 0 → x = −5 or x = 3.
For what value of k does kx² + 12x + 9 = 0 have exactly one real solution?
Show solution
One solution ⇒ discriminant = 0: 12² − 4(k)(9) = 0.
144 = 36k → k = 4.
The Northside Method — How We Teach This 1-on-1
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