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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

x^2+2x+2x+4-162=0

We add all the numbers together, and all the variables

x^2+4x-158=0

a = 1; b = 4; c = -158;
Δ = b2-4ac
Δ = 42-4·1·(-158)
Δ = 648
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-18\sqrt{2}}{2*1}=\frac{-4-18\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+18\sqrt{2}}{2*1}=\frac{-4+18\sqrt{2}}{2}
$