Solve the equation (x+5)2 = 7(x−5) for x.

The left side is $(x+5)^2$, which expands to $x^2 + 2(x)(5) + 5^2$.
The right side is $7(x-5)$, which expands to $7x - 7(5)$.

$$x^2 + 10x + 25 = 7x - 35$$

Subtract $7x$ from both sides and add 35 to both sides.

$$x^2 + 10x - 7x + 25 + 35 = 0$$ $$x^2 + 3x + 60 = 0$$

For the equation $x^2 + 3x + 60 = 0$, we have $a=1$, $b=3$, and $c=60$.

$$x = \frac{-(3) \pm \sqrt{(3)^2 - 4(1)(60)}}{2(1)}$$ $$x = \frac{-3 \pm \sqrt{9 - 240}}{2}$$ $$x = \frac{-3 \pm \sqrt{-231}}{2}$$

Since the discriminant $\Delta = -231$ is negative, the solutions will be complex numbers. We can write $\sqrt{-231}$ as $i\sqrt{231}$, where $i = \sqrt{-1}$.

$$x = \frac{-3 \pm i\sqrt{231}}{2}$$

The solutions are:

$$x_1 = \frac{-3 + i\sqrt{231}}{2}$$ $$x_2 = \frac{-3 - i\sqrt{231}}{2}$$