Solve $LaTeX: \displaystyle x^{2} + 42\geq 14 x - 3$

Getting zero on the right hand side and factoring gives $LaTeX: \displaystyle x^{2} - 14 x + 45=0\iff \left(x - 9\right) \left(x - 5\right)=0$ . This gives the critical numbers as $LaTeX: \displaystyle x=9$ and $LaTeX: \displaystyle x=5$ .

Checking the sign in each region gives the solution $LaTeX: \displaystyle (-\infty, 5 ] \cup [ 9 ,\infty)$ .