(x+5)(3x-4)-3(x+2)+4=0

Solution for (x+5)(3x-4)-3(x+2)+4=0 equation:

(x+5)(3x-4)-3(x+2)+4=0

We multiply parentheses

(x+5)(3x-4)-3x-6+4=0

We multiply parentheses ..

(+3x^2-4x+15x-20)-3x-6+4=0

We add all the numbers together, and all the variables

(+3x^2-4x+15x-20)-3x-2=0

We get rid of parentheses

3x^2-4x+15x-3x-20-2=0

We add all the numbers together, and all the variables

3x^2+8x-22=0

a = 3; b = 8; c = -22;
Δ = b2-4ac
Δ = 82-4·3·(-22)
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{82}}{2*3}=\frac{-8-2\sqrt{82}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{82}}{2*3}=\frac{-8+2\sqrt{82}}{6}
$