We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where the discriminant is non-negative, i.e., \(\geq 0\),
Let us consider the following quadratic equation:
\(a x^2+b x+c=0\) with real coefficients \(a, b, c\) and \(a \neq 0\).
Also, let us assume that the \(b^2-4 a c<0\).
Now, we know that we can find the square root of negative real numbers in the set of complex numbers. Therefore, the solutions to the above equation are available in the set of complex numbers which are given by
\(
x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-b \pm \sqrt{4 a c-b^2} i}{2 a}
\)

“A polynomial equation has at least one root.”
As a consequence of this theorem, the following result, which is of immense importance, is arrived at:
“A polynomial equation of degree \(n\) has \(n\) roots.”

Proof of Quadratic formula

\(a x^2+b x+c=0\) is a standard algebraic form of a quadratic equation in mathematics. Let’s derive the quadratic formula from this equation.

\(
\begin{aligned}
&a x^2+b x+c=0\\
&x^2+\frac{b}{a} x+\frac{c}{a}=0\\
&x^2+\frac{b}{a} x=-\frac{c}{a}\\
&x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2 a}\right)^2\\
&\left(x+\frac{b}{2 a}\right)^2=-\frac{c}{a}+\frac{b^2}{4 a^2}=\frac{b^2-4 a c}{4 a^2}\\
&x+\frac{b}{2 a}=\pm \sqrt{\frac{b^2-4 a c}{4 a^2}}=\pm \frac{\sqrt{b^2-4 a c}}{2 a}\\
&x=-\frac{b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}
\end{aligned}
\)

Example 1: Solve \(x^2+x+1=0\)

Solution: Here, \(\quad b^2-4 a c=1^2-4 \times 1 \times 1=1-4=-3\)
\(\text { Therefore, the solutions are given by } x=\frac{-1 \pm \sqrt{-3}}{2 \times 1}=\frac{-1 \pm \sqrt{3} i}{2}\)

Example 2: Solve \(\sqrt{5} x^2+x+\sqrt{5}=0\)

Solution: Here, the discriminant of the equation is
\(
1^2-4 \times \sqrt{5} \times \sqrt{5}=1-20=-19
\)
Therefore, the solutions are
\(
\frac{-1 \pm \sqrt{-19}}{2 \sqrt{5}}=\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}} .
\)