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)= (-b\pm i √(-\Delta) )/(2a) = (8 \pm 10i)/(2) \\ \\ X1=4-5i \\ \\ X2=4+5i`