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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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