(11x+8)(x-7)+(3x+2)(x-7)=0

Solution for (11x+8)(x-7)+(3x+2)(x-7)=0 equation:

(11x+8)(x-7)+(3x+2)(x-7)=0

We multiply parentheses ..

(+11x^2-77x+8x-56)+(3x+2)(x-7)=0

We get rid of parentheses

11x^2-77x+8x+(3x+2)(x-7)-56=0

We multiply parentheses ..

11x^2+(+3x^2-21x+2x-14)-77x+8x-56=0

We add all the numbers together, and all the variables

11x^2+(+3x^2-21x+2x-14)-69x-56=0

We get rid of parentheses

11x^2+3x^2-21x+2x-69x-14-56=0

We add all the numbers together, and all the variables

14x^2-88x-70=0

a = 14; b = -88; c = -70;
Δ = b2-4ac
Δ = -882-4·14·(-70)
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-108}{2*14}=\frac{-20}{28}
=-5/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+108}{2*14}=\frac{196}{28}
=7
$