You solve this question pls

Step 1: Apply the Pythagorean theorem. 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 $x+8$. $x^2 + (x+7)^2 = (x+8)^2$

Step 2: Expand the squared terms. $x^2 + (x^2 + 14x + 49) = (x^2 + 16x + 64)$

Step 3: Simplify the equation. $2x^2 + 14x + 49 = x^2 + 16x + 64$

Step 4: Rearrange the equation to form a standard quadratic equation ($ax^2 + bx + c = 0$). Subtract $x^2$, $16x$, and $64$ from both sides: $2x^2 - x^2 + 14x - 16x + 49 - 64 = 0$ $x^2 - 2x - 15 = 0$

Step 5: Solve the quadratic equation by factoring. We need two numbers that multiply to $-15$ and add to $-2$. These numbers are $-5$ and $3$. $(x-5)(x+3) = 0$

Step 6: Find the possible values for $x$. $x-5 = 0 \implies x = 5$ $x+3 = 0 \implies x = -3$

Step 7: Choose the valid solution. Since $x$ represents a length, it must be a positive value. Therefore, $x=-3$ is not a valid solution. $x = 5$

Step 8: Calculate the lengths of the sides. Side 1: $x = 5$ Side 2: $x+7 = 5+7 = 12$ Side 3: $x+8 = 5+8 = 13$

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