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.

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

$x-2=0" "rarr x =2$

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