1. We are asked to solve the quadratic equation $m^2 = 4(2m + 5)$. 2. First, expand the right side: $$m^2 = 8m + 20$$ 3. Rearrange all terms to one side to set the equation to zero: $$m^2 - 8m - 20 = 0$$ 4. This is a quadratic equation in standard form $ax^2 + bx + c = 0$ where $a=1$, $b=-8$, and $c=-20$. 5. Use the quadratic formula to solve for $m$: $$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 6. Substitute the values: $$m = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-20)}}{2(1)} = \frac{8 \pm \sqrt{64 + 80}}{2} = \frac{8 \pm \sqrt{144}}{2}$$ 7. Simplify the square root: $$\sqrt{144} = 12$$ 8. Find the two solutions: $$m = \frac{8 + 12}{2} = \frac{20}{2} = 10$$ $$m = \frac{8 - 12}{2} = \frac{-4}{2} = -2$$ 9. Therefore, the solutions to the equation are $m = 10$ and $m = -2$.