Step 1 :

After factoring out $ 4 $ we have:

$$ 28x^{2}+16x-80 = 4 ( 7x^{2}+4x-20 ) $$

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:

$$ a = 7 , b = 4 ~ \text{ and } ~ c = -20 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = -20} $.

$$ a \cdot c = -140 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -140 $ and add to $ b = 4 $.

Step 5: All pairs of numbers with a product of $ -140 $ are:

PRODUCT = -140
-1     140	1     -140
-2     70	2     -70
-4     35	4     -35
-5     28	5     -28
-7     20	7     -20
-10     14	10     -14

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 4 }$

PRODUCT = -140 and SUM = 4
-1     140	1     -140
-2     70	2     -70
-4     35	4     -35
-5     28	5     -28
-7     20	7     -20
-10     14	10     -14

Step 7: Replace middle term $ 4 x $ with $ 14x-10x $:

$$ 7x^{2}+4x-20 = 7x^{2}+14x-10x-20 $$

Step 8: Apply factoring by grouping. Factor $ 7x $ out of the first two terms and $ -10 $ out of the last two terms.

$$ 7x^{2}+14x-10x-20 = 7x\left(x+2\right) -10\left(x+2\right) = \left(7x-10\right) \left(x+2\right) $$

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