What is the solution of the equation (x – 5)2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.

Answer:x = 1/5 (25 + 3 i sqrt(5)) or x = (-3 i sqrt(5) + 25)/5Step-by-step explanation using the quadratic formula:Solve for x: 5 (x - 5)^2 + 9 = 0 Expand out terms of the left hand side: 5 x^2 - 50 x + 134 = 0 x = (50 ± sqrt((-50)^2 - 4×5×134))/(2×5) = (50 ± sqrt(2500 - 2680))/10 = (50 ± sqrt(-180))/10: x = (50 + sqrt(-180))/10 or x = (50 - sqrt(-180))/10 sqrt(-180) = sqrt(-1) sqrt(180) = i sqrt(180): x = (50 + i sqrt(180))/10 or x = (50 - i sqrt(180))/10 sqrt(180) = sqrt(4×9×5) = sqrt(2^2×3^2×5) = 2×3sqrt(5) = 6 sqrt(5): x = (i×6 sqrt(5) + 50)/10 or x = (-i×6 sqrt(5) + 50)/10 Factor 2 from 50 + 6 i sqrt(5) giving 2 (3 i sqrt(5) + 25): x = 1/102 (3 i sqrt(5) + 25) or x = (-6 i sqrt(5) + 50)/10 (2 (3 i sqrt(5) + 25))/10 = (2 (3 i sqrt(5) + 25))/(2×5) = (3 i sqrt(5) + 25)/5: x = (3 i sqrt(5) + 25)/5 or x = (-6 i sqrt(5) + 50)/10 Factor 2 from 50 - 6 i sqrt(5) giving 2 (-3 i sqrt(5) + 25): x = 1/5 (25 + 3 i sqrt(5)) or x = 1/102 (-3 i sqrt(5) + 25) (2 (-3 i sqrt(5) + 25))/10 = (2 (-3 i sqrt(5) + 25))/(2×5) = (-3 i sqrt(5) + 25)/5: Answer: x = 1/5 (25 + 3 i sqrt(5)) or x = (-3 i sqrt(5) + 25)/5