How do you solve the quadratic #4x^2-12x=16# using any method?
2 Answers
Sep 18, 2016
#x=4#
#x=-1#
Explanation:
Given -
#4x^2-12x=16#
#4x^2-12x-16=0#
#4x^2+4x-16x-16=0#
#4x(x+1)-16(x+1)=0#
#(4x-16)(x+1)=0#
#4x-16=0#
#x=16/4=4#
#x+1=0#
#x=-1#
Sep 18, 2016
Explanation:
Find factors of 4 which subtract to give 3.
If
If