(x+1)(x-7)+(x+1)(3x-2)=0

Solution for (x+1)(x-7)+(x+1)(3x-2)=0 equation:

(x+1)(x-7)+(x+1)(3x-2)=0

We multiply parentheses ..

(+x^2-7x+x-7)+(x+1)(3x-2)=0

We get rid of parentheses

x^2-7x+x+(x+1)(3x-2)-7=0

We multiply parentheses ..

x^2+(+3x^2-2x+3x-2)-7x+x-7=0

We add all the numbers together, and all the variables

x^2+(+3x^2-2x+3x-2)-6x-7=0

We get rid of parentheses

x^2+3x^2-2x+3x-6x-2-7=0

We add all the numbers together, and all the variables

4x^2-5x-9=0

a = 4; b = -5; c = -9;
Δ = b2-4ac
Δ = -52-4·4·(-9)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*4}=\frac{-8}{8}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*4}=\frac{18}{8}
=2+1/4
$