How do you solve $x^2+6x-16=0$ ?

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Answer 1 · 2016-03-27T00:06:26

$x=2$ or $x=-8$

Find two factors of $16$ which differ by $6$ . The pair $8, 2$ works in that $8 xx 2 = 16$ and $8 - 2 = 6$ .

Hence we find:

$0 = x^2+6x-16 = (x+8)(x-2)$

So $x = 2$ or $x = -8$

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Alternatively, complete the square then use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

with $a = (x+3)$ and $b=5$ as follows:

$0 = x^2+6x-16$

$=(x+3)^2-9-16$

$=(x+3)^2-25$

$=(x+3)^2-5^2$

$=((x+3)-5)((x+3)+5)$

$=(x-2)(x+8)$

Hence zeros $x=2$ and $x=-8$

How do you solve x^2+6x-16=0x^2+6x-16=0?