x + 2 * sqrt(2x + 5) = 8

Subtract $x$ from both sides of the equation.

$$2\sqrt{2x+5} = 8 - x$$

For the square root to be defined, $2x+5 \ge 0 \Rightarrow x \ge -\frac{5}{2}$.
Also, for the right side to be equal to a non-negative square root, $8-x \ge 0 \Rightarrow x \le 8$.
So, any valid solution must satisfy $-\frac{5}{2} \le x \le 8$.

$$(2\sqrt{2x+5})^2 = (8 - x)^2$$ $$4(2x+5) = 64 - 16x + x^2$$ $$8x + 20 = 64 - 16x + x^2$$

Move all terms to one side.

$$0 = x^2 - 16x - 8x + 64 - 20$$ $$0 = x^2 - 24x + 44$$

We can factor the quadratic equation. We need two numbers that multiply to 44 and add up to -24. These numbers are -2 and -22.

$$(x - 2)(x - 22) = 0$$

This gives two potential solutions:

$$x - 2 = 0 \Rightarrow x = 2$$ $$x - 22 = 0 \Rightarrow x = 22$$

Check $x = 2$:

$$2 + 2\sqrt{2(2)+5} = 8$$ $$2 + 2\sqrt{4+5} = 8$$ $$2 + 2\sqrt{9} = 8$$ $$2 + 2(3) = 8$$ $$2 + 6 = 8$$ $$8 = 8$$

This solution is valid and satisfies the condition $-\frac{5}{2} \le x \le 8$.

Check $x = 22$:

$$22 + 2\sqrt{2(22)+5} = 8$$ $$22 + 2\sqrt{44+5} = 8$$ $$22 + 2\sqrt{49} = 8$$ $$22 + 2(7) = 8$$ $$22 + 14 = 8$$ $$36 = 8$$

This statement is false, so $x=22$ is an extraneous solution. It also does not satisfy the condition $x \le 8$.

The only valid solution is $x=2$.