5x(4x-3)=0

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

5x(4x-3)=0

We multiply parentheses

20x^2-15x=0

a = 20; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·20·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*20}=\frac{30}{40}
=3/4
$