Solve for x in the equation x2-8x+41=0

Answer:d) 4 ± 5iStep-by-step explanation:Here we have to use the quadratic formula.x = [tex]\frac{-b +/- \sqrt{b^2 - 4ac} }{2a}[/tex]In the given equation x^2 - 8x + 41 = 0, a =1, b = -8 and c = 41Now plug in the given values in the above formula, we getx = [tex]\frac{-(-8) +/- \sqrt{(-8)^2 - 4*1*41} }{2*1}[/tex]Simplifying the above, we getx = [tex]\frac{8 +/- \sqrt{64 - 164} }{2}[/tex]x = [tex]\frac{8 +/- \sqrt{-100} }{2}[/tex][√-100 = √-1 *√100 = i*10 = 10i] because the value of √-1 = i]x = (8 ± 10i )/2Now dividing by 2, we getx = 4 ± 5iThe answer is d) 4 ± 5iHope you will understand the concept.Thank you.