Solve it! Is it easy?

I (taking the easy way out) tested 8 and 20 in the equation rather than solving it algebraically. 8 workes and 20 did not, so the answer is 8.

The question can be expressed as
$\begin{aligned} & \sqrt {2x + 9} = 13 - x\\ & \implies (2x+9) = {(13-x)}^2\\ & \implies(2x+9) = x^2 - 26x + 169\\ & \implies x^2 -28x+160 = 0\\ & \implies x^2 - 20x - 8x + 160 = 0\\ & \implies x(x-20) -8(x-20) = 0\\ & \implies (x-20)(x-8) = 0\\ & \implies x-20 = 0 \text {(or)} x-8=0\\ & \implies x = 20 \text {(or)} x = 8 \end{aligned}$

But, if we substitute $x = 20$ in the question, the equation does not satisfy, so $x$ is not equal to $20$ .

So, $x = \boxed {8}$

First thing to note is that 2x+9 is under root, so it can't be negative. So 2x+9>0. So x>-9/2.And Root 2x+9 Is positive so it should be less than 13. Squaring both sides and then solving the equation we find the value of x to be 8 or 20. But x can't be greater than 13. So the value of x is 8.