We are given the inequality $\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2$ and the condition that $m \neq -\frac{5}{2}$. We need to simplify the inequality.

1. Problem Description

We are given the inequality $\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2$ and the condition that $m \neq -\frac{5}{2}$. We need to simplify the inequality.

2. Solution Steps

First, combine the fractions on the left side of the inequality:

$\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2$

$\frac{-(m^2 - 20)}{2m+5} - \frac{m^2 + m - 12}{2m+5} < 2$

$\frac{-m^2 + 20 - m^2 - m + 12}{2m+5} < 2$

$\frac{-2m^2 - m + 32}{2m+5} < 2$

Now, subtract 2 from both sides:

$\frac{-2m^2 - m + 32}{2m+5} - 2 < 0$

$\frac{-2m^2 - m + 32 - 2(2m+5)}{2m+5} < 0$

$\frac{-2m^2 - m + 32 - 4m - 10}{2m+5} < 0$

$\frac{-2m^2 - 5m + 22}{2m+5} < 0$

$\frac{-(2m^2 + 5m - 22)}{2m+5} < 0$

$\frac{2m^2 + 5m - 22}{2m+5} > 0$

Let's find the roots of the quadratic equation $2m^2 + 5m - 22 = 0$. Using the quadratic formula,

$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$m = \frac{-5 \pm \sqrt{5^2 - 4(2)(-22)}}{2(2)}$

$m = \frac{-5 \pm \sqrt{25 + 176}}{4}$

$m = \frac{-5 \pm \sqrt{201}}{4}$

So the roots are $m_1 = \frac{-5 - \sqrt{201}}{4} \approx -4.79$ and $m_2 = \frac{-5 + \sqrt{201}}{4} \approx 2.29$.

Also, $2m+5 = 0$ when $m = -\frac{5}{2} = -2.5$.

Now, we analyze the sign of $\frac{2m^2 + 5m - 22}{2m+5}$. We have three critical points: $m_1 \approx -4.79$, $m = -2.5$ and $m_2 \approx 2.29$.

We test the sign of the expression in the intervals $(-\infty, -4.79)$, $(-4.79, -2.5)$, $(-2.5, 2.29)$, and $(2.29, \infty)$.

1. $m < -4.79$, let $m = -5$.

$\frac{2(-5)^2 + 5(-5) - 22}{2(-5)+5} = \frac{50 - 25 - 22}{-10+5} = \frac{3}{-5} < 0$.

2. $-4.79 < m < -2.5$, let $m = -3$.

$\frac{2(-3)^2 + 5(-3) - 22}{2(-3)+5} = \frac{18 - 15 - 22}{-6+5} = \frac{-19}{-1} > 0$.

3. $-2.5 < m < 2.29$, let $m = 0$.

$\frac{2(0)^2 + 5(0) - 22}{2(0)+5} = \frac{-22}{5} < 0$.

4. $m > 2.29$, let $m = 3$.

$\frac{2(3)^2 + 5(3) - 22}{2(3)+5} = \frac{18 + 15 - 22}{6+5} = \frac{11}{11} > 0$.

Since we want $\frac{2m^2 + 5m - 22}{2m+5} > 0$, the solution is $(-4.79, -2.5) \cup (2.29, \infty)$, or $(\frac{-5 - \sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5 + \sqrt{201}}{4}, \infty)$.

3. Final Answer

The solution to the inequality $\frac{m^2 - 20}{-(2m+5)} - \frac{m^2 + m - 12}{2m+5} < 2$ is $m \in (\frac{-5 - \sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5 + \sqrt{201}}{4}, \infty)$.

Final Answer: $m \in (\frac{-5-\sqrt{201}}{4}, -\frac{5}{2}) \cup (\frac{-5+\sqrt{201}}{4}, \infty)$