solving this inequality?

solving this inequality?

\(-3x^2-12 \le 3x^2-5x-68\)

-3x^2-12 \le 3x^2-5x-68

\(\begin{array}{|rcll|} \hline -3x^2-12 &\le& 3x^2-5x-68 \\ -3x^2-12 -(3x^2-5x-68 ) &\le& 0 \\ -6x^2 + 5x + 56 &\le& 0 \\ \hline \end{array}\)

Now all we have to do is find the zeroes of \( y = -6x^2 + 5x + 56\)

\(\begin{array}{|rcll|} \hline -6x^2 + 5x + 56 &=& 0 \\ x &=& \frac{-5\pm \sqrt{25-4\cdot(-6)\cdot 56 } }{2\cdot(-6)} \\ x &=& \frac{-5\pm \sqrt{1369 } }{-12} \\ x &=& \frac{-5\pm 37 }{-12} \\\\ x_1 &=& \frac{-5+37}{-12} \\ x_1 &=& \frac{32}{-12} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{-\frac{8}{3}} \\\\ x_2 &=& \frac{-5-37}{-12} \\ x_2 &=& \frac{-42}{-12} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{\frac{7}{2}} \\ \hline \end{array}\)

Since \(y = -6x^2 + 5x + 56\)  graphs as a down parabola,
the quadratic is below the axis on the ends:

Then the solution is:
    \(x \le -\frac{8}{3} \quad \text{ or } \quad x \ge \frac{7}{2}\)