Trigonometry 7th Edition

Comparing the equation with the general form $ax^{2}+bx+c$, we get a=1, b= -4 and c=13. $b^{2}-4ac$ is less than 0. Therefore, the solutions are given by $x= \frac{-b±\sqrt {4ac-b^{2}}i}{2a}=\frac{4±\sqrt {4\times1\times13-16}i}{2}$ or $x=2+3i$ or $x=2-3i$. Hence proved.