Discriminant sign determines the root count

Every step of this proof reuses the identity $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$ established in the previous theorem, which holds for any real $a,b,c$ with $a\neq0$ regardless of the sign of $\Delta$ — only the final square-root step depended on $\Delta\ge0$.

**Case $\Delta>0$.** The right side $\dfrac{\Delta}{4a^2}$ is a positive number (a positive numerator over a positive denominator $4a^2$), so it has two distinct square roots, $+\dfrac{\sqrt{\Delta}}{2a}$ and $-\dfrac{\sqrt{\Delta}}{2a}$, which differ because $\sqrt{\Delta}\neq0$. Each gives a different value of $x+\dfrac{b}{2a}$, hence a different value of $x$: exactly two distinct real roots.

**Case $\Delta=0$.** The right side becomes $0$, and the only real number whose square is $0$ is $0$ itself — there is no "$\pm$" ambiguity left. So $x+\dfrac{b}{2a}=0$, giving the single root $x=-\dfrac{b}{2a}$. This matches factoring the left side directly as $a\left(x+\dfrac{b}{2a}\right)^2=0$, a perfect square touching zero exactly once.

**Case $\Delta<0$.** The right side $\dfrac{\Delta}{4a^2}$ is now negative. But the left side $\left(x+\dfrac{b}{2a}\right)^2$ is a square of a real number, and the square of any real number is always $\ge0$ — it can never equal a negative number. So no real $x$ can satisfy $\left(x+\dfrac{b}{2a}\right)^2=\dfrac{\Delta}{4a^2}$ in this case, meaning the original equation has no real solution at all.