2x(x+1)+8x=(x+3)(x+5)

Solution for 2x(x+1)+8x=(x+3)(x+5) equation:

2x(x+1)+8x=(x+3)(x+5)

We move all terms to the left:

2x(x+1)+8x-((x+3)(x+5))=0

We add all the numbers together, and all the variables

8x+2x(x+1)-((x+3)(x+5))=0

We multiply parentheses

2x^2+8x+2x-((x+3)(x+5))=0

We multiply parentheses ..

2x^2-((+x^2+5x+3x+15))+8x+2x=0


We calculate terms in parentheses: -((+x^2+5x+3x+15)), so:

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

We get rid of parentheses

x^2+5x+3x+15

We add all the numbers together, and all the variables

x^2+8x+15

Back to the equation:

-(x^2+8x+15)


We add all the numbers together, and all the variables

2x^2+10x-(x^2+8x+15)=0

We get rid of parentheses

2x^2-x^2+10x-8x-15=0

We add all the numbers together, and all the variables

x^2+2x-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$