Question

x + 3 26 10

Explanation:

Step1: Apply Pythagorean theorem

$$(x + 3)^2+10^2=26^2$$

Step2: Expand and simplify

$$x^2+6x + 9+100=676$$
$$x^2+6x+109 = 676$$
$$x^2+6x-567 = 0$$

Step3: Solve quadratic equation

For $ax^2+bx + c = 0$ ($a = 1$, $b = 6$, $c=-567$), use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$$x=\frac{-6\pm\sqrt{6^2-4\times1\times(-567)}}{2\times1}=\frac{-6\pm\sqrt{36 + 2268}}{2}=\frac{-6\pm\sqrt{2304}}{2}=\frac{-6\pm48}{2}$$
We take the positive root (since length cannot be negative)
$$x=\frac{-6 + 48}{2}=\frac{42}{2}=21$$

Answer:

$x = 21$