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 ?

Rembember that $(a-b)^2 = a^2 + b^2 -2ab$ which isn't $a^2 - b^2$.  So $\sqrt{7-x} - \sqrt{x-13}$ squared is
$$
(7-x) + (x-13) - 2\sqrt{(7-x)(x-13)}
$$
Now you can isolate the square root term and square again.
