How do you solve the equation #2x^2-7x+12=0# by completing the square? Precalculus Linear and Quadratic Functions Completing the Square 1 Answer Binayaka C. May 11, 2018 Solution: # x = 1.75 +- 1/4 sqrt(47) i # Explanation: #2 x^2 -7 x +12 =0 or 2 x^2 -7 x = -12 # or # 2 (x^2 -7/2 x )= -12 # or # 2 (x^2 -7/2 x +49/16 )= 49/8-12 # or # 2 (x -7/4)^2 = -47/8 # or # (x -7/4)^2 = -47/16 # or # (x -7/4)= +- sqrt(-47/16) # or # (x -7/4)= +- sqrt(47i^2)/4 [i^2=-1] # # x = 7/4+- 1/4sqrt(47) i # or # x = 1.75 +- 1/4sqrt(47) i # Solution: # x = 1.75 +- 1/4 sqrt(47) i # [Ans] Answer link Related questions What does completing the square mean? How do I complete the square? Does completing the square always work? Is completing the square always the best method? Do I need to complete the square if #f(x) = x^2 - 6x + 9#? How do I complete the square if #f(x) = x^2 + 4x - 9#? How do I complete the square if the coefficient of #x^2# is not 1? How do I complete the square if #f(x) = 3x^2 + 12x - 9#? If I know the quadratic formula, why must I also know how to complete the square? How do I use completing the square to describe the graph of #f(x)=30-12x-x^2#? See all questions in Completing the Square Impact of this question 1437 views around the world You can reuse this answer Creative Commons License