How do you solve $x^2+8x-12=0$ ?

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Answer 1 · 2016-12-12T01:35:54

$x=-2-2sqrt7$ or $x=-2+2sqrt7$

We can solve $x^2+8x-12=0$ by using completing square method, as follows.

$x^2+8x-12=0$ can be written as

$x^2+2×4×x+4^2-4^2-12=0$

or $(x+2)^2-16-12=0$

or $(x+2)^2-28=0$

or $(x+2)^2-(sqrt28)^2=0$

and as factors of $a^2-b^2$ are $(a+b)$ and $(a-b)$ , quadratic polynomial on LHS can be factorized as under:

$(x+2+sqrt28)(x+2-sqrt28)=0$

and hence either $x+2+sqrt28=0$ i.e. $x=-2-sqrt28=-2-2sqrt7$

or $x+2-sqrt28=0$ i.e. $x=-2+sqrt28=-2+2sqrt7$

How do you solve x^2+8x-12=0x^2+8x-12=0?