162=(x+2)(2x+4)

Solution for 162=(x+2)(2x+4) equation:

162=(x+2)(2x+4)

We move all terms to the left:

162-((x+2)(2x+4))=0

We multiply parentheses ..

-((+2x^2+4x+4x+8))+162=0


We calculate terms in parentheses: -((+2x^2+4x+4x+8)), so:

(+2x^2+4x+4x+8)

We get rid of parentheses

2x^2+4x+4x+8

We add all the numbers together, and all the variables

2x^2+8x+8

Back to the equation:

-(2x^2+8x+8)


We get rid of parentheses

-2x^2-8x-8+162=0

We add all the numbers together, and all the variables

-2x^2-8x+154=0

a = -2; b = -8; c = +154;
Δ = b2-4ac
Δ = -82-4·(-2)·154
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-36}{2*-2}=\frac{-28}{-4}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+36}{2*-2}=\frac{44}{-4}
=-11
$