Solve $LaTeX: \displaystyle x^{2} - 45\geq - 3 x - 5$

Getting zero on the right hand side and factoring gives $LaTeX: \displaystyle x^{2} + 3 x - 40=0\iff \left(x - 5\right) \left(x + 8\right)=0$ . This gives the critical numbers as $LaTeX: \displaystyle x=5$ and $LaTeX: \displaystyle x=-8$ .

Checking the sign in each region gives the solution $LaTeX: \displaystyle (-\infty, -8 ] \cup [ 5 ,\infty)$ .