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

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

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

We multiply parentheses ..

(+2x^2+x+14x+7)+(x+7)(x+5)=0

We get rid of parentheses

2x^2+x+14x+(x+7)(x+5)+7=0

We multiply parentheses ..

2x^2+(+x^2+5x+7x+35)+x+14x+7=0

We add all the numbers together, and all the variables

2x^2+(+x^2+5x+7x+35)+15x+7=0

We get rid of parentheses

2x^2+x^2+5x+7x+15x+35+7=0

We add all the numbers together, and all the variables

3x^2+27x+42=0

a = 3; b = 27; c = +42;
Δ = b2-4ac
Δ = 272-4·3·42
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-15}{2*3}=\frac{-42}{6}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+15}{2*3}=\frac{-12}{6}
=-2
$