Step 1 :

After factoring out $ 2 $ we have:

$$ 14x^{2}-74x+20 = 2 ( 7x^{2}-37x+10 ) $$

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:

$$ a = 7 , b = -37 ~ \text{ and } ~ c = 10 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 10} $.

$$ a \cdot c = 70 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 70 $ and add to $ b = -37 $.

Step 5: All pairs of numbers with a product of $ 70 $ are:

PRODUCT = 70
1     70	-1     -70
2     35	-2     -35
5     14	-5     -14
7     10	-7     -10

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -37 }$

PRODUCT = 70 and SUM = -37
1     70	-1     -70
2     35	-2     -35
5     14	-5     -14
7     10	-7     -10

Step 7: Replace middle term $ -37 x $ with $ -2x-35x $:

$$ 7x^{2}-37x+10 = 7x^{2}-2x-35x+10 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 7x^{2}-2x-35x+10 = x\left(7x-2\right) -5\left(7x-2\right) = \left(x-5\right) \left(7x-2\right) $$

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