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Solve for $x$ in the equation $−x^2−3x+5\geq 0$

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$-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2}$

Explanation

$x=\frac{3+\sqrt{29}}{-2}, \frac{3-\sqrt{29}}{-2}$
(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,
$-\frac{3+\sqrt{29}}{2} \leq x \leq-\frac{3-\sqrt{29}}{2}$

by Platinum (122,714 points)

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