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

1 Answer
Nov 16, 2016

Answer:

#x=7/4+-(isqrt(-47))/4#

Explanation:

Given -

#2x^2-7x+12=0#

#2x^2-7x=-12#
Divide both sides by #2#
#(cancel(2)x^2)/cancel(2)-7/2x=(-12)/2#
#x^2-7/2x=-6#
Divide# 7/2# by #2#, square it and add it to both sides

#x^2-7/2*1/2*x +49/16=-6+49/16#
#x^2-7/4+49/16=-6+49/16#
#x^2-7/4+49/16=(-96+49)/16#
#(x-7/4)^2=(-47)/16#
#x-7/4=+-sqrt((-47)/16)#
#x=7/4+-(isqrt(-47))/4#