The Quadratic Formula Solves Every Quadratic Equation

Factoring is efficient when a quadratic has recognizable factors. The quadratic formula works for every equation $ax^2+bx+c=0$ with $a\ne0$ because it is derived from equivalence-preserving algebra.

Derive the formula

Divide by $a$:

$x^2+\frac ba x=-\frac ca.$

Add $(b/2a)^2$ to both sides:

$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$

Taking both square-root branches and isolating $x$ gives

The $\pm$ is essential because both positive and negative numbers can square to the same value.

The discriminant predicts the roots

Let $D=b^2-4ac$.

• $D>0$: two distinct real roots;
• $D=0$: one repeated real root;
• $D<0$: two complex-conjugate roots and no real horizontal intercepts.

Read structure from the formula

The average of the two roots is $-b/(2a)$, exactly the axis of symmetry. Their separation is $\sqrt D/|a|$. The formula therefore encodes the geometry of the parabola, not only a solving procedure.

Check your understanding

Without solving, classify the roots of $3x^2+2x+5=0$.

Show the reasoning

$D=2^2-4(3)(5)=4-60=-56<0$. There are two complex-conjugate roots and no real intercepts.