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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-20x^2+15x-16x+12)=0

We get rid of parentheses

-20x^2+15x-16x+12=0

We add all the numbers together, and all the variables

-20x^2-1x+12=0

a = -20; b = -1; c = +12;
Δ = b2-4ac
Δ = -12-4·(-20)·12
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-31}{2*-20}=\frac{-30}{-40}
=3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+31}{2*-20}=\frac{32}{-40}
=-4/5
$