ĐK: $x^2-5x-6 \geq 0 \implies \left[\begin{matrix} x \leq -1 & \quad \\ x \geq 6& \quad \end{matrix}\right.$ $(1')$
$\sqrt{x^2-5x-6} + 2x^2 > 10x + 15$
$\implies \sqrt{x^2-5x-6}> -2x^2 + 10x + 15 \,\, (1)$
TH1: $-2x^2 + 10x + 15 < 0 \implies \left[\begin{matrix} x < \dfrac{5-\sqrt{55}}{2} & \quad \\ x > \dfrac{5+\sqrt{55}}{2} & \quad \end{matrix}\right.$ (Thõa mãn điều kiện ban đầu)
Khi đó $(1)$ luôn đúng với $\forall x \in \left(-\infty; \dfrac{5-\sqrt{55}}{2}\right) \cup \left(\dfrac{5+\sqrt{55}}{2}; +\infty \right) \,\, (1'')$
TH2: $-2x^2 + 10x + 15 \geq 0 \implies \dfrac{5-\sqrt{55}}{2} \leq x \leq \dfrac{5+\sqrt{55}}{2} \,\, (2')$
Khi đó:
$(1) \implies x^2-5x-6 > (-2x^2 + 10x + 15)^2$
$\implies x^2-5x-6 > 4x^4 - 40x^3 + 40x^2 + 300x + 225$
$\implies 4x^4 - 40x^3 + 39x^2 + 305x + 231 < 0$
$\implies (4x^2 - 20x - 33)(x^2 - 5x - 7) < 0$
$\implies \left[ \begin{array}{l}
\left\{
\begin{array}{1}
4x^2 - 20x - 33 > 0\\
x^2-5x-7 < 0\\
\end{array}
\right.
\\
\left\{
\begin{array}{1}
4x^2 - 20x - 33 < 0\\
x^2-5x-7 > 0\\
\end{array}
\right.
\end{array} \right.$
$\implies \left[ \begin{array}{l}
\left\{ \begin{array}{1} x \in \left(-\infty; \dfrac{5-\sqrt{58}}{2}\right) \cup \left(\dfrac{5+\sqrt{58}}{2}; +\infty \right) \\ \dfrac{5-\sqrt{53}}{2} < x < \dfrac{5+\sqrt{53}}{2}\\ \end{array} \right.
\\
\left\{ \begin{array}{1} \dfrac{5-\sqrt{58}}{2} < x < \dfrac{5+\sqrt{58}}{2} \\ x \in \left(-\infty; \dfrac{5-\sqrt{53}}{2}\right) \cup \left(\dfrac{5+\sqrt{53}}{2}; +\infty \right) \\ \end{array} \right.
\end{array} \right.$
$\implies \left[ \begin{array}{l}
x \in \varnothing
\\
x \in \left(\dfrac{5-\sqrt{58}}{2}; \dfrac{5-\sqrt{53}}{2}\right) \cup \left(\dfrac{5+\sqrt{53}}{2}; \dfrac{5+\sqrt{58}}{2} \right)
\end{array} \right.$
Kết hợp với $(1')$ và $(2')$ $\implies x \in \left[\dfrac{5-\sqrt{55}}{2}; \dfrac{5-\sqrt{53}}{2}\right) \cup \left(\dfrac{5+\sqrt{53}}{2}; \dfrac{5+\sqrt{55}}{2} \right] \,\, (2'')$
Từ $(1'')$ và $(2'')$
$\implies x \in R$ \ $\left[\dfrac{5-\sqrt{53}}{2}; \dfrac{5+\sqrt{53}}{2}\right]$