(x+5)(2x+1)-(x+5)(x-7)=0

Solution for (x+5)(2x+1)-(x+5)(x-7)=0 equation:

(x+5)(2x+1)-(x+5)(x-7)=0

We multiply parentheses ..

(+2x^2+x+10x+5)-(x+5)(x-7)=0

We get rid of parentheses

2x^2+x+10x-(x+5)(x-7)+5=0

We multiply parentheses ..

2x^2-(+x^2-7x+5x-35)+x+10x+5=0

We add all the numbers together, and all the variables

2x^2-(+x^2-7x+5x-35)+11x+5=0

We get rid of parentheses

2x^2-x^2+7x-5x+11x+35+5=0

We add all the numbers together, and all the variables

x^2+13x+40=0

a = 1; b = 13; c = +40;
Δ = b2-4ac
Δ = 132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*1}=\frac{-10}{2}
=-5
$