How do you complete the square to solve $x^2+2x-10=14$ ?

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Answer 1 · 2015-05-23T20:26:45

If you first add $11$ to both sides you get:

$x^2+2x+1 = 25$

Now

$x^2+2x+1 = (x+1)(x+1) = (x+1)^2$

and $25 = 5^2$

So we have

$(x+1)^2 = 5^2$

Hence

$x+1 = +-sqrt(5^2) = +-5$

Subtracting 1 from both sides we get:

$x = -1 +-5$

That is $x = 4$ or $x=-6$

How do you complete the square to solve x^2+2x-10=14x^2+2x-10=14?