Square root equation I have the equation $\sqrt{(7-x)} - \sqrt {(x+13)} = 2 $ The square root should be expanded so it is square root of $7-x$ - square root of $x+13 = 2$. 
When i square both sides i get:
$7-x - x-13 = 4 $
then i clean on the LHS so i get
$-2x-6 = 4$ which leads to $(-2x-6)^2 = 4^2$ and after working that out i get:
$4x^2 + 24 x + 36 = 16$.
Next step will then be $4x^2 +24x +20 = 0$
Using the pq formula i get $x^2 = -6/2$ +- $\sqrt {6 \over 2}^2 -5$
And after that i get $x^2 = -6/2 +- 2$ which leads to $x^2 = -1$ or $x^2= -5$.
The answer according to the textbook is $-9$. Do you know how to solve this ?

You probably mean

$$
\sqrt{7-x} - \sqrt{x+13} = 2.
$$

The steps:

$$
\big( \sqrt{7-x} - \sqrt{x+13} \big)^2 = 2^2\\
\Downarrow\\
\sqrt{7-x}^2 + \sqrt{x+13}^2 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
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\big(7-x\big) + \big(x+13\big) - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
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20 - 2 \sqrt{7-x} \sqrt{x+13} = 4\\
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\sqrt{7-x} \sqrt{x+13} = 8\\
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\big(7-x\big)\big(x+13\big) = 64\\
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\big(10-3-x\big)\big(10+3+x\big) = 64\\
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100-\big(3+x\big)^2 = 64\\
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\big(3+x\big)^2 = 36\\
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3+x = \pm 6\\
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\bbox[16px,border:2px solid #800000] {x = -9 \vee x = 3}
$$

