How do you solve #4x^2+6x=12# by completing the square?

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Answer 1 · 2017-03-21T12:40:15

The solutions are #S={-2.637, 1.137}#

Before completing the square, we divide by #2#

#4x^2+6x=12#

#2x^2+3x=6#

#2(x^2+3/2x)=6#

#(x^2+3/2x)=3#

Now, we complete the square

#(x^2+3/2x+9/16)=3+9/16#

#(x+3/4)^2=57/16#

#x+3/4=+-sqrt57/16#

#x=-3/4+-sqrt57/16#

#x=1/4(-3+-sqrt57)#

Therefore,

#x_1=1/4(-3-sqrt57)=-2.637#

#x_2=1/4(-3+sqrt57)=1.137#