Answer
The solution of the equation ${{x}^{2}}-2x+17=0$ in standard form is $\left\{ 1+4i,1-4i \right\}$.
Work Step by Step
Consider the equation,${{x}^{2}}-2x+17=0$
Compare the equation ${{x}^{2}}-2x+17=0$ with $a{{x}^{2}}+bx+c$.
$\begin{align}
& a=1 \\
& b=-2 \\
& c=17
\end{align}$
Substitute $a=1$, $b=-2$ and $c=17$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
& x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 17 \right)}}{2\left( 1 \right)} \\
& =\frac{2\pm \sqrt{4-68}}{2} \\
& =\frac{2\pm \sqrt{-64}}{2}
\end{align}$
Use the property $\sqrt{-b}=i\sqrt{b}$.
\[\begin{align}
& x=\frac{2\pm i\sqrt{64}}{2} \\
& =\frac{2\pm 8i}{2} \\
& =\frac{2}{2}\pm \frac{8}{2}i \\
& =1\pm 4i
\end{align}\]
Therefore, the solution of the equation ${{x}^{2}}-2x+17=0$ in standard form is $\left\{ 1+4i,1-4i \right\}$.
