One formula that solves every quadratic, and a quantity that forecasts the roots in advance.
Completing the square works on every quadratic. So here's a lazy, brilliant idea: instead of completing the square over and over for each new equation, let's complete it once — on the general equation ax2 + bx + c = 0 — and keep the answer forever. What drops out is a master key called the quadratic formula. Feed it the three numbers a, b, c and out come the roots. And the little expression living under its square root quietly forecasts, before you finish a single calculation, whether you'll get two roots, one, or none.
By the end you'll be able to derive the formula, plug numbers into it without tripping over signs, compute the discriminant Δ = b2 − 4ac to predict the roots, and even work backward — choosing a missing number so the equation has exactly the roots you want. Throughout, the colors carry the meaning: a is blue (the x² term), b is teal (the x term), c is amber (the constant), and every root is violet. The discriminant Δ gets its own red call-out, because it's the part most worth watching.
The formula isn't magic and it isn't something to memorize on faith. It's just completing the square done once, with letters instead of numbers, so that the work is banked for all time. Let's walk the whole derivation slowly. We start with the general quadratic, and we'll assume a ≠ 0 (otherwise there's no x² term and it isn't a quadratic at all).
The one move that makes completing the square possible is getting a 1 in front of x². So our very first step is to divide every term by a. After that, completing the square means taking half of the x-coefficient and squaring it — that's the famous (b⁄2a)2 — and adding it to both sides to keep the equation balanced. The left side then folds up into a perfect square, the right side gets combined over a common denominator, and a single square root finishes the job.
That last line is the whole point. It says: to solve any quadratic, you never have to complete the square again. Just read off a, b, c and substitute. The two signs in the ± give you the two roots — one from the plus, one from the minus.
Tap "Next step" to reveal the derivation one honest line at a time. The final tap boxes the master key.
Using the formula is a four-step ritual: (1) write the equation as ax2 + bx + c = 0, (2) read off a, b, c with their signs, (3) compute the thing under the root, b2 − 4ac, and (4) drop everything into (−b ± √…)/(2a). The single most common slip is forgetting that it's −b, not b. When b is already negative, −b turns positive — and people miss that flip constantly.
Solve 2x2 − 7x + 3 = 0. Here a = 2, b = −7, c = 3.
Solve x2 + 2x − 1 = 0. Here a = 1, b = 2, c = −1.
That's the formula's superpower over factoring: it doesn't care whether the roots are whole numbers, fractions, or ugly irrationals. Factoring only works smoothly when the answer happens to be nice. The formula works always.
Set a, b, c and watch Δ and the two roots compute live, with the parabola dropping violet dots on its x-intercepts.
Look again at the formula. Everything is fixed arithmetic except the part under the square root, b2 − 4ac. That expression has its own name: the discriminant, written with the Greek letter delta, Δ. It "discriminates" — it tells the roots apart.
Here's why it's so useful. You can compute Δ all by itself, in one line, before you do any of the rest. And just from its sign you already know the whole story of the roots — without solving. It's like checking the weather forecast before you leave the house: you don't need to step outside to know whether to grab an umbrella.
For ax2 + bx + c = 0, the discriminant is Δ = b2 − 4ac. It is the quantity living under the square root in the quadratic formula, and its sign alone decides how many real roots there are.
Why does the sign of Δ control everything? Because the formula takes √Δ. Think about what that square root does in three cases.
If Δ > 0, then √Δ is a real positive number, and the ± splits it into two genuinely different answers — the −b + √Δ branch and the −b − √Δ branch land in different places. Two distinct real roots. The parabola crosses the x-axis at two points.
If Δ = 0, then √Δ = 0, and adding or subtracting zero gives the same thing both times. The two roots merge into one repeated root, sitting exactly at x = −b⁄(2a). The parabola just touches the x-axis at its vertex — a kiss, not a crossing.
If Δ < 0, you'd be taking the square root of a negative number, which has no real value. No real roots. The parabola floats entirely above or below the x-axis and never reaches it.
The discriminant tells you how many real roots, not what they are. A positive Δ that isn't a perfect square (like Δ = 8) still means two real roots — they're just irrational. And Δ being negative doesn't mean "no answer ever"; it means no real answer. Those cases reopen later as complex numbers.
Keep x2 − 2x + c and slide c. Watch the parabola cut, touch, or clear the x-axis as Δ changes sign.
Once you trust that Δ's sign controls the roots, you can run the logic backward. Someone tells you what kind of roots they want, you set up the matching condition on Δ, and you solve for the unknown number in the equation. This is a favorite test question, and it's the gateway to inequalities.
Suppose x2 − 6x + k = 0 is supposed to have two equal roots (a repeated root). "Equal roots" is exactly the Δ = 0 case. So set Δ = (−6)2 − 4·1·k = 36 − 4k equal to zero: 36 − 4k = 0 → k = 9. So k = 9 is the one value that makes the parabola just kiss the axis.
Now suppose x2 + 4x + k = 0 must have two distinct real roots. That's Δ > 0, an inequality: 16 − 4k > 0 → 16 > 4k → k < 4. Notice the answer is a whole range of values, not a single number. And "no real roots" would be Δ < 0, giving k > 4. This move — translating "kind of roots" into an inequality on a parameter — is precisely the doorway into Stage 12.
Slide k in x2 − 6x + k = 0. Find the single value of k that makes Δ = 0 and the roots merge.
Complete the square once on the general quadratic and you get a master key, x = (−b ± √(b2 − 4ac))⁄(2a), that solves any ax2 + bx + c = 0 by substitution — remembering it's −b and the whole numerator sits over 2a. The expression under the root is the discriminant Δ = b2 − 4ac, and its sign forecasts the roots: Δ > 0 two, Δ = 0 one, Δ < 0 none. Read that logic backward and "what kind of roots?" becomes an equation or inequality you solve for a missing parameter.
We've been finding where parabolas hit the x-axis. Next we turn the picture around and study the parabola itself — its vertex, its axis of symmetry, and how a, b, c shape it — so that solving and graphing become two views of one object.
Work each one out first, then open the answer to check your thinking.
Tap the answer you think is right.
This lesson covers Common Core A-REI.B.4b: solving quadratics by the quadratic formula, and using the discriminant to reason about the number and nature of solutions. The misconception to watch hardest for is the sign of the leading term in the numerator — students routinely write b instead of −b, and lose the flip when b is itself negative — plus the parenthesization that keeps the whole numerator over 2a. Deriving the formula by completing the square (rather than presenting it as a fact) is deliberate: it shows the formula is earned, and it ties this lesson directly to 11.2. The Δ < 0 / "no real roots" case is where complex numbers (Stage 31) begin, and the backward "find k so that…" problems with their inequality answers are exactly where Stage 12 (quadratic inequalities) takes root.