Step 1 :

After factoring out $ 3 $ we have:

$$ 60x^{2}-69x+18 = 3 ( 20x^{2}-23x+6 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 20 , b = -23 ~ \text{ and } ~ c = 6 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 20 }$ by the constant term $\color{blue}{c = 6} $.

$$ a \cdot c = 120 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 120 $ and add to $ b = -23 $.

Step 5: All pairs of numbers with a product of $ 120 $ are:

PRODUCT = 120
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -23 }$

PRODUCT = 120 and SUM = -23
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 7: Replace middle term $ -23 x $ with $ -8x-15x $:

$$ 20x^{2}-23x+6 = 20x^{2}-8x-15x+6 $$

Step 8: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -3 $ out of the last two terms.

$$ 20x^{2}-8x-15x+6 = 4x\left(5x-2\right) -3\left(5x-2\right) = \left(4x-3\right) \left(5x-2\right) $$

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