How do you solve #x^2+4x-12=0#?

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Answer 1 · 2016-11-17T08:08:32

#(x+6)(x-2) =0# gives

#x=2" or "x = -6#

#x^2 +4x -12 = 0#

Factorise this quadratic trinomial first.

Find factors of 12 which subtract (because of the MINUS) to make 4.

The correct factors will be #6 and 2#

Their signs must be DIFFERENT (because of the MINUS 12)

The greater must be positive (because of the PLUS 4)

#(x+6)(x-2) =0#

Now either of the two factors can be 0, to give the product of 0.

#x+6=0 " "rarr x =-6#

#x-2=0" "rarr x =2#