If $b^2-4ac\geq 0$, then the root of quadratic equation $ax^2+bx+c=0$ is

$\rightarrow$ Given Quadratic equation is $a x^{2}+b x+c=0$ and also

$\Delta=b^{2}-4 a c \geqslant 0$

$\therefore$ Roots are real and distinct.

let the roots be $\alpha, \beta$.

we know that

$\begin{array}{c} \alpha+\beta=\frac{-b}{a}……(1) \\ \alpha \beta=\frac{c}{a} . \end{array}$

$\begin{array}{c} \Rightarrow(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=\frac{b^{2}}{a^{2}}-\frac{4 c}{a}=\frac{b^{2}-4 a c}{a^{2}} \\ \Rightarrow \alpha-\beta= \pm \frac{\sqrt{b^{\prime \prime}-4 a c}}{a} \longrightarrow \text { (2) } \end{array}$

(1) +(2)

$\begin{array}{l} \Rightarrow 2 \alpha=\frac{-b \pm \sqrt{b^{2}-4 a c}}{a} \\ \Rightarrow \alpha=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a .} \end{array}$

$\therefore$ root of the Quadratic equation is $\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$.