(x+3)(x+1)=2x^2-18

Solution for (x+3)(x+1)=2x^2-18 equation:

(x+3)(x+1)=2x^2-18

We move all terms to the left:

(x+3)(x+1)-(2x^2-18)=0

We get rid of parentheses

-2x^2+(x+3)(x+1)+18=0

We multiply parentheses ..

-2x^2+(+x^2+x+3x+3)+18=0

We get rid of parentheses

-2x^2+x^2+x+3x+3+18=0

We add all the numbers together, and all the variables

-1x^2+4x+21=0

a = -1; b = 4; c = +21;
Δ = b2-4ac
Δ = 42-4·(-1)·21
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10}{2*-1}=\frac{-14}{-2}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10}{2*-1}=\frac{6}{-2}
=-3
$