Có bao nhiêu giá trị nguyên của tham số $a\in \left[ -10;10...

Xét hàm số $h\left( x \right)=3{{x}^{4}}-4\left( a+2 \right){{x}^{3}}+12a{{x}^{2}}-30a$
$\Rightarrow {h}'\left( x \right)=12{{x}^{3}}-12\left( a+2 \right){{x}^{2}}+24ax$
Trường hợp 1:
$\begin{aligned}
& \left\{ \begin{aligned}
& {h}'\left( x \right)\le 0,\forall \in \left( -\infty ;-2 \right) \\
& h\left( -2 \right)\ge 0 \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-2x-ax+2a\ge 0 \\
& 112+50a\ge 0 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& x\left( x-2 \right)\ge a\left( x-2 \right) \\
& a\ge -2,24 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& a\ge x,\forall x\in \left( -\infty ;-2 \right) \\
& a\ge -2,24 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\ge -2 \\
& a\ge -2,24 \\
\end{aligned} \right. \\
& \Leftrightarrow a\ge -2. \\
\end{aligned}$
Trường hợp 2:
$\begin{aligned}
& \left\{ \begin{aligned}
& {h}'\left( x \right)\ge 0,\forall \in \left( -\infty ;-2 \right) \\
& h\left( -2 \right)\le 0 \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-2x-ax+2a\le 0 \\
& 112+50a\le 0 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& x\left( x-2 \right)\le a\left( x-2 \right) \\
& a\le -2,24 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a\le x,\forall x\in \left( -\infty ;-2 \right) \\
& a\ge -2,24 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& a\in \varnothing \\
& a\ge -2,24 \\
\end{aligned} \right.\Leftrightarrow a\in \varnothing . \\
\end{aligned}$
Vậy $a\in \left[ -2;10 \right]$ nên có $13$ giá trị nguyên $a$ thỏa bài toán.