Solve $LaTeX: \displaystyle x^{2} + 71\geq 18 x - 1$

Getting zero on the right hand side and factoring gives $LaTeX: \displaystyle x^{2} - 18 x + 72=0\iff \left(x - 12\right) \left(x - 6\right)=0$ . This gives the critical numbers as $LaTeX: \displaystyle x=12$ and $LaTeX: \displaystyle x=6$ .

Checking the sign in each region gives the solution $LaTeX: \displaystyle (-\infty, 6 ] \cup [ 12 ,\infty)$ .