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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -6; b = -2; c = +20;
Δ = b2-4ac
Δ = -22-4·(-6)·20
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*-6}=\frac{-20}{-12}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*-6}=\frac{24}{-12}
=-2
$