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

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

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

We multiply parentheses ..

(+15x^2+12x-25x-20)=0

We get rid of parentheses

15x^2+12x-25x-20=0

We add all the numbers together, and all the variables

15x^2-13x-20=0

a = 15; b = -13; c = -20;
Δ = b2-4ac
Δ = -132-4·15·(-20)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-37}{2*15}=\frac{-24}{30}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+37}{2*15}=\frac{50}{30}
=1+2/3
$