How do you solve #2x^2-7x+12=0# by completing the square?
1 Answer
Nov 16, 2016
#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#
