#### Answer

$$[1-\sqrt{3},1+\sqrt{3}]$$

#### Work Step by Step

Let $x^2 \leq 2x +2 $ be a function $f$.
Express the inequality in the form $f(x) \lt 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$x^2 \leq 2x +2 $
$x^2 -2x -2\leq 2x +2-2x -2$
$x^2 -2x -2\leq 0 $
Find the $x$-intercepts by solving $x^2 -2x -2=0$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-2$, $c=-2$
$x = \frac{-(-2)±\sqrt{-2^2-4⋅1⋅-2}}{2⋅1}$
$x = 1+ \sqrt{3}$ or $1- \sqrt{3}$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, 1- \sqrt{3})(1- \sqrt{3},1+ \sqrt{3})(1+ \sqrt{3}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, 1- \sqrt{3})$
Test Value = $-1$
$f=x^2-2x-2$
$f=-1^2-2(-1)-2$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(-∞,1-\sqrt{3})$.
$(1-\sqrt{3}, 1+sqrt{3})$
Test Value = $0$
$f=x^2-2x-2$
$f=0^2-2(0)-2$
$f=-2$
Conclusion: $f (x) < 0$ for all $x$ in $(1-\sqrt{3}, 1+\sqrt{3})$.
$(1+\sqrt{3},+∞)$
Test Value = $3$
$f=x^2-2x-2$
$f=3^2-2(3)-2$
$f=1$
Conclusion: $f (x) > 0$ for all $x$ in $(1+\sqrt{3},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \leq 0$, where $f (x) = x^2-2x-2$. Based on the solution above, $f(x)\lt0$ for all $x$ in $(1-\sqrt{3},1+\sqrt{3})$. However, because the inequality involves $\leq$ (less than or equal to), we must also include the solutions of $x^2-2x-2$, namely $1-\sqrt{3}$ and $1+\sqrt{3}$, in the solution set.
Thus, the solution set of the given inequality is:
$$[1-\sqrt{3},1+\sqrt{3}]$$