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

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Answer 1 · 2016-12-12T01:35:54

$x = - 2 - 2 \sqrt{7}$ $x=-2-2sqrt7$ or $x = - 2 + 2 \sqrt{7}$ $x=-2+2sqrt7$

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

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

$x^{2} + 2 \times 4 \times x + 4^{2} - 4^{2} - 12 = 0$ $x^2+2×4×x+4^2-4^2-12=0$

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

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

or ${(x + 2)}^{2} - {(\sqrt{28})}^{2} = 0$ $(x+2)^2-(sqrt28)^2=0$

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

$(x + 2 + \sqrt{28}) (x + 2 - \sqrt{28}) = 0$ $(x+2+sqrt28)(x+2-sqrt28)=0$

and hence either $x + 2 + \sqrt{28} = 0$ $x+2+sqrt28=0$ i.e. $x = - 2 - \sqrt{28} = - 2 - 2 \sqrt{7}$ $x=-2-sqrt28=-2-2sqrt7$

or $x + 2 - \sqrt{28} = 0$ $x+2-sqrt28=0$ i.e. $x = - 2 + \sqrt{28} = - 2 + 2 \sqrt{7}$ $x=-2+sqrt28=-2+2sqrt7$

How do you solve x2+8x−12=0x^2+8x-12=0?