Step 1 :

After factoring out $ 2 $ we have:

$$ 6x^{2}-20x+14 = 2 ( 3x^{2}-10x+7 ) $$

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:

$$ a = 3 , b = -10 ~ \text{ and } ~ c = 7 $$

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

$$ a \cdot c = 21 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 21 $ and add to $ b = -10 $.

Step 5: All pairs of numbers with a product of $ 21 $ are:

PRODUCT = 21
1     21	-1     -21
3     7	-3     -7

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -10 }$

PRODUCT = 21 and SUM = -10
1     21	-1     -21
3     7	-3     -7

Step 7: Replace middle term $ -10 x $ with $ -3x-7x $:

$$ 3x^{2}-10x+7 = 3x^{2}-3x-7x+7 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 3x^{2}-3x-7x+7 = 3x\left(x-1\right) -7\left(x-1\right) = \left(3x-7\right) \left(x-1\right) $$

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