(3x+5)(2x-4)-20=0

Solution for (3x+5)(2x-4)-20=0 equation:

(3x+5)(2x-4)-20=0

We multiply parentheses ..

(+6x^2-12x+10x-20)-20=0

We get rid of parentheses

6x^2-12x+10x-20-20=0

We add all the numbers together, and all the variables

6x^2-2x-40=0

a = 6; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·6·(-40)
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{241}}{2*6}=\frac{2-2\sqrt{241}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{241}}{2*6}=\frac{2+2\sqrt{241}}{12}
$