Expand.
\[2{x}^{2}+3x-2x-3\le 02322\]
Simplify \(2{x}^{2}+3x-2x-3\) to \(2{x}^{2}+x-3\).
\[2{x}^{2}+x-3\le 02322\]
Move all terms to one side.
\[2{x}^{2}+x-3-02322\le 0\]
Simplify \(2{x}^{2}+x-3-02322\) to \(2{x}^{2}+x-2325\).
\[2{x}^{2}+x-2325\le 0\]
Use the Quadratic Formula.
In general, given \(a{x}^{2}+bx+c=0\), there exists two solutions where:
\[x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}\]
In this case, \(a=2\), \(b=1\) and \(c=-2325\).
\[{x}^{}=\frac{-1+\sqrt{1-4\times 2\times -2325}}{2\times 2},\frac{-1-\sqrt{1-4\times 2\times -2325}}{2\times 2}\]
Simplify.
\[x=\frac{-1+\sqrt{18601}}{4},\frac{-1-\sqrt{18601}}{4}\]
\[x=\frac{-1+\sqrt{18601}}{4},\frac{-1-\sqrt{18601}}{4}\]
From the values of \(x\) above, we have these 3 intervals to test.
\[\begin{aligned}&x\le \frac{-1-\sqrt{18601}}{4}\\&\frac{-1-\sqrt{18601}}{4}\le x\le \frac{-1+\sqrt{18601}}{4}\\&x\ge \frac{-1+\sqrt{18601}}{4}\end{aligned}\]
Pick a test point for each interval.
For the interval \(x\le \frac{-1-\sqrt{18601}}{4}\):
Let's pick \(x=-35\). Then, \((-35-1)(2\times -35+3)\le 02322\).After simplifying, we get \(2412\le 1234\), which is
false
.
Drop this interval.
.
For the interval \(\frac{-1-\sqrt{18601}}{4}\le x\le \frac{-1+\sqrt{18601}}{4}\):
Let's pick \(x=0\). Then, \((0-1)(2\times 0+3)\le 02322\).After simplifying, we get \(-3\le 1234\), which is
true
.
Keep this interval.
.
For the interval \(x\ge \frac{-1+\sqrt{18601}}{4}\):
Let's pick \(x=34\). Then, \((34-1)(2\times 34+3)\le 02322\).After simplifying, we get \(2343\le 1234\), which is
false
.
Drop this interval.
.
Therefore,
\[\frac{-1-\sqrt{18601}}{4}\le x\le \frac{-1+\sqrt{18601}}{4}\]
(-1-sqrt(18601))/4<=x<=(-1+sqrt(18601))/4
