How do you factor $-x^{2}-12x-18=0$ ?

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Answer 1 · 2018-04-22T14:29:37

The solution is $x=-6±3sqrt2$

and hence the factorization is
$(x+6-3sqrt2)(x+6+3sqrt2))=0$

$-x^2-12x-18=0$
$x^2+12x+18=0$

$x= (-b+-sqrt(b^2-4ac))/(2a)$

$x= (-12+-sqrt(144-72))/(2 times1)$

$x=(-12+-sqrt72)/2$

$x= (-12+-6sqrt2)/2$

$x=-6±3sqrt2$

These are the solutions when $-x^2-12x-18=0$

The factorization can be written in the form $(x-a)(x-b)$ where $a$ and $b$ are the solutions, also known as zeros.

$(x-(-6+3sqrt2))(x-(-6-3sqrt2))=0$

$(x+6-3sqrt2)(x+6+3sqrt2))=0$

How do you factor -x^{2}-12x-18=0-x^{2}-12x-18=0?