The quadratic formula & the discriminant

Run completing-the-square once on the general quadratic ax² + bx + c = 0 — divide by a, move c, add the corner, root, tidy — and the answer arrives as a formula that works for every quadratic, forever:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Nothing in it is arbitrary: the −b and the 2a are the "move and divide" steps, and the square root holds the completed corner. It's the last lesson's geometry, frozen into symbols.

Using it

Solve 2x² + 5x − 3 = 0. Read off a = 2, b = 5, c = −3:

$x = \frac{-5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}$

So x = 1/2 or x = −3. Two habits keep the formula honest: write the equation in ax² + bx + c = 0 form first (a stray term on the right corrupts a, b, c), and carry signs with their coefficients (here c = −3, and −4ac became +24).

The discriminant: the part that predicts

The expression under the root, D = b² − 4ac, decides everything before you finish:

D < 0 leaves a negative number under the square root — impossible for real numbers, which is precisely why the floating parabola has no x-intercepts. (Algebra II's complex numbers give even those quadratics solutions; the formula survives intact.)

Choosing your weapon

Why this matters

This is the course's summit: equations (module 1) + expanding and factoring (modules 3–4) + perfect squares, all condensed into one always-works formula. From here the story continues in Pre-Calculus, where parabolas join the wider family of functions — transformed, composed, and inverted.