Question

Solve x^2 – 8x + 41 = 0 for x.

Answer 1

$\displaystyle x^2 - 8x + 41 = 0 \\\\ \Delta=b^2-4ac \\ \Delta=(-8)^2-4\cdot1\cdot41 \\ \Delta=64-164 \\ \Delta=-100 \\ \\ X_{1,2}= \frac{-b\pm i \sqrt{-\Delta} }{2a} = \frac{8 \pm 10i}{2} \\ \\ X1=4-5i \\ \\ X2=4+5i$