You solve this question pls

In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
The sides are $x$, $x+7$, and $x+8$. The hypotenuse is $x+8$.

$$x^2 + (x+7)^2 = (x+8)^2$$

Subtract $x^2$, $16x$, and 64 from both sides:

$$2x^2 - x^2 + 14x - 16x + 49 - 64 = 0$$ $$x^2 - 2x - 15 = 0$$

We need two numbers that multiply to $-15$ and add to $-2$. These numbers are $-5$ and 3.

$$(x-5)(x+3) = 0$$

$$x-5 = 0 \implies x = 5$$ $$x+3 = 0 \implies x = -3$$

Since $x$ represents a length, it must be a positive value. Therefore, $x=-3$ is not a valid solution.

$$x = 5$$

Side 1: $x = 5$
Side 2: $x+7 = 5+7 = 12$
Side 3: $x+8 = 5+8 = 13$

The value of $x$ is $\boxed{5}$.