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

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

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

We multiply parentheses ..

(+20x^2-12x+15x-9)=0

We get rid of parentheses

20x^2-12x+15x-9=0

We add all the numbers together, and all the variables

20x^2+3x-9=0

a = 20; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·20·(-9)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-27}{2*20}=\frac{-30}{40}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+27}{2*20}=\frac{24}{40}
=3/5
$