Compute E(4X−5X^2)

Note $$4X - 5X^2 = 500 - 96X - 5(X-10)^2,$$ so that $$\operatorname{E}[4X - 5X^2] = \operatorname{E}[500] - 96 \operatorname{E}[X] - 5 \operatorname{E}[(X-10)^2],$$ and since $\operatorname{E}[X] = 10$ and $\operatorname{Var}[X] = \operatorname{E}[(X-10)^2] = 20$, it follows that

$$\operatorname{E}[4X-5X^2] = 500 - 960 - 100 = -560.$$

Note that we did not need to use the fact that $X$ is gamma distributed.

From $\text{Var}(X)=E(X^2)-(E(X))^2$ you can compute $E(X^2)$ to be $\text{Var}(X)+100 = 120$ indeed.

The rest is just applying linearity of expectation. $-$ instead of $+$ is no problem there. I mean, you expect the difference of two dice to be $0$, right?