\[3{x}^{2}+2x>5\]
1
Move all terms to one side.
3x2+2x−5>03{x}^{2}+2x-5>0

2
Split the second term in 3x2+2x−53{x}^{2}+2x-5 into two terms.
3x2+5x−3x−5>03{x}^{2}+5x-3x-5>0

3
Factor out common terms in the first two terms, then in the last two terms.
x(3x+5)−(3x+5)>0x(3x+5)-(3x+5)>0

4
Factor out the common term 3x+53x+5.
(3x+5)(x−1)>0(3x+5)(x-1)>0

5
Solve for xx.
x=−53,1x=-\frac{5}{3},1

6
From the values of xx above, we have these 3 intervals to test.
x<−53−53<x<1x>1\begin{aligned}&x<-\frac{5}{3}\\&-\frac{5}{3}<x<1\\&x>1\end{aligned}

7
Pick a test point for each interval.
For the interval x<−53x<-\frac{5}{3}:

Let's pick x=−2x=-2. Then, 3(−2)2+2×−2>53{(-2)}^{2}+2\times -2>5.
After simplifying, we get 8>58>5, which is
true
.
Keep this interval.
.

For the interval −53<x<1-\frac{5}{3}<x<1:

Let's pick x=0x=0. Then, 3×02+2×0>53\times {0}^{2}+2\times 0>5.
After simplifying, we get 0>50>5, which is
false
.
Drop this interval.
.

For the interval x>1x>1:

Let's pick x=2x=2. Then, 3×22+2×2>53\times {2}^{2}+2\times 2>5.
After simplifying, we get 16>516>5, which is
true
.
Keep this interval.
.

8
Therefore,
x<−53,x>1x<-\frac{5}{3},x>1

Done
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