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

1 Answer
May 11, 2018

Answer:

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]