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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6x^2+16x-21x-56)+11x)=0


We calculate terms in parentheses: -((+6x^2+16x-21x-56)+11x), so:

(+6x^2+16x-21x-56)+11x

We get rid of parentheses

6x^2+16x-21x+11x-56

We add all the numbers together, and all the variables

6x^2+6x-56

Back to the equation:

-(6x^2+6x-56)


We get rid of parentheses

-6x^2-6x+56=0

a = -6; b = -6; c = +56;
Δ = b2-4ac
Δ = -62-4·(-6)·56
Δ = 1380
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1380}=\sqrt{4*345}=\sqrt{4}*\sqrt{345}=2\sqrt{345}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{345}}{2*-6}=\frac{6-2\sqrt{345}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{345}}{2*-6}=\frac{6+2\sqrt{345}}{-12}
$