(x+1)(3x-5)-(2x+4)(x+1)=0

Solution for (x+1)(3x-5)-(2x+4)(x+1)=0 equation:

(x+1)(3x-5)-(2x+4)(x+1)=0

We multiply parentheses ..

(+3x^2-5x+3x-5)-(2x+4)(x+1)=0

We get rid of parentheses

3x^2-5x+3x-(2x+4)(x+1)-5=0

We multiply parentheses ..

3x^2-(+2x^2+2x+4x+4)-5x+3x-5=0

We add all the numbers together, and all the variables

3x^2-(+2x^2+2x+4x+4)-2x-5=0

We get rid of parentheses

3x^2-2x^2-2x-4x-2x-4-5=0

We add all the numbers together, and all the variables

x^2-8x-9=0

a = 1; b = -8; c = -9;
Δ = b2-4ac
Δ = -82-4·1·(-9)
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{-(-8)-10}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+10}{2*1}=\frac{18}{2}
=9
$