Step 1 :

After factoring out $ -3 $ we have:

$$ -60x^{2}+21x+18 = -3 ( 20x^{2}-7x-6 ) $$

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:

$$ a = 20 , b = -7 ~ \text{ and } ~ c = -6 $$

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 = -7 $.

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 $\color{blue}{ b = -7 }$

PRODUCT = -120 and SUM = -7
-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 $ -7 x $ with $ 8x-15x $:

$$ 20x^{2}-7x-6 = 20x^{2}+8x-15x-6 $$

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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