## Precalculus (6th Edition) Blitzer

The solution of the equation is $-1+2i\,\,\,or\,\,\,-1-2i$.
Comparing the provided equation with the general quadratic equation, we get $a=1,b=2\,\,\text{ and }\,\,c=5$ Substitute the value of $a,b\,\,\text{ and }\,\,c$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, to get \begin{align} & x=\frac{-2\pm \sqrt{{{2}^{2}}-4\cdot 1\cdot 5}}{2\cdot 1} \\ & =\frac{-2\pm \sqrt{4-20}}{2} \\ & =\frac{-2\pm 4i}{2} \\ & =-1\pm 2i \\ & =-1+2i\,\,\,or\,\,\,-1-2i \end{align} Therefore, the solution of the equation is $-1+2i\,\,\,or\,\,\,-1-2i$.