Answer
The solution set for the quadratic equation, ${{x}^{2}}-2x+4=0$ is $\left\{ 1\pm i\sqrt{3} \right\}$.
Work Step by Step
Consider the quadratic equation,
${{x}^{2}}-2x+4=0$
Compare the provided equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$.
Here, $a=1,\text{ }b=-2\text{ and }c=4$
Apply the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute 1 for a, $-2$ for b and 4 for c.
$x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}$
Simplify the radical.
$\begin{align}
& x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)} \\
& =\frac{2\pm \sqrt{4-16}}{2} \\
& =\frac{2\pm \sqrt{-12}}{2}
\end{align}$
As $i=\sqrt{-1}$
$\begin{align}
& x=\frac{2\pm \sqrt{12}\sqrt{-1}}{2} \\
& =\frac{2\pm 2i\sqrt{3}}{2} \\
& =\frac{2\left( 1\pm i\sqrt{3} \right)}{2} \\
& =1\pm i\sqrt{3}
\end{align}$
The solutions of the quadratic are complex conjugates of each other.
Hence, the solution set for the quadratic equation ${{x}^{2}}-2x+4=0$ is $\left\{ 1\pm i\sqrt{3} \right\}$.
