Answer
\[
-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2}
\]
Explanation
(1) Use the Quadratic Formula.
\[
x=\frac{3+\sqrt{29}}{-2}, \frac{3-\sqrt{29}}{-2}
\]
(2) From the values of \(x\) above, we have these 3 intervals to test.
\[
\begin{aligned}
&x \leq-\frac{3+\sqrt{29}}{2} \\
&-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2} \\
&x \geq-\frac{3-\sqrt{29}}{2}
\end{aligned}
\]
(3) Pick a test point for each interval.
For the interval \(x \leq-\frac{3+\sqrt{29}}{2}\) :
Let's pick \(x=-5\). Then, \(-(-5)^{2}-3 \times-5+5 \geq 0\).
After simplifying, we get \(-5 \geq 0\), which is false.
Drop this interval..
For the interval \(-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2}\) :
Let's pick \(x=0\). Then, \(-0^{2}-3 \times 0+5 \geq 0\).
After simplifying, we get \(5 \geq 0\), which is true.
Keep this interval..
For the interval \(x \geq-\frac{3-\sqrt{29}}{2}\) :
Let's pick \(x=2\). Then, \(-2^{2}-3 \times 2+5 \geq 0\).
After simplifying, we get \(-5 \geq 0\), which is false.
Drop this interval..
(4) Therefore,
\[
-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2}
\]
