How do you solve the equation #2x^2-7x+12=0# by completing the square?

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Answer 1 · 2018-05-11T12:34:20

Solution: # x = 1.75 +- 1/4 sqrt(47) i #

#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]