Answer
Solution set: $ [1-\sqrt{3},1+\sqrt{3}]$
Work Step by Step
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^{2} \leq 2x+2$
$x^{2}-2x-2 \leq 0$
... no integer factors of $-2$ that add up to $-2$...
... no factoring, then ...
... use the quadratic formula in the next step...
$f(x)=x^{2}-2x-2 \leq 0$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$x^{2}-4x+2 = 0$
$x=\displaystyle \frac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}$
$=\displaystyle \frac{4\pm\sqrt{12}}{2}=\frac{2(1\pm\sqrt{3})}{2}=1\pm\sqrt{3}$
3. Locate these boundary points on a number line, thereby dividing the number line into intervals.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & (-\infty, 1-\sqrt{3}) & (1-\sqrt{3},1+\sqrt{3}) & (1+\sqrt{3},\infty)\\
a=test.val. & -10 & 1 & 10\\
f(a) & 100+20-2 & 1-2-2 & 100-40-2\\
f(a) \leq 0 ? & F & T & F
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $ [1-\sqrt{3},1+\sqrt{3}]$