(x+3)(x+1)=2(x+3)(x-5)

Solution for (x+3)(x+1)=2(x+3)(x-5) equation:

(x+3)(x+1)=2(x+3)(x-5)

We move all terms to the left:

(x+3)(x+1)-(2(x+3)(x-5))=0

We multiply parentheses ..

(+x^2+x+3x+3)-(2(x+3)(x-5))=0


We calculate terms in parentheses: -(2(x+3)(x-5)), so:

2(x+3)(x-5)

We multiply parentheses ..

2(+x^2-5x+3x-15)

We multiply parentheses

2x^2-10x+6x-30

We add all the numbers together, and all the variables

2x^2-4x-30

Back to the equation:

-(2x^2-4x-30)


We get rid of parentheses

x^2-2x^2+x+3x+4x+3+30=0

We add all the numbers together, and all the variables

-1x^2+8x+33=0

a = -1; b = 8; c = +33;
Δ = b2-4ac
Δ = 82-4·(-1)·33
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-14}{2*-1}=\frac{-22}{-2}
=+11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+14}{2*-1}=\frac{6}{-2}
=-3
$