5(4r-1)(r+3)=0

Solution for 5(4r-1)(r+3)=0 equation:

5(4r-1)(r+3)=0

We multiply parentheses ..

5(+4r^2+12r-1r-3)=0

We multiply parentheses

20r^2+60r-5r-15=0

We add all the numbers together, and all the variables

20r^2+55r-15=0

a = 20; b = 55; c = -15;
Δ = b2-4ac
Δ = 552-4·20·(-15)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-65}{2*20}=\frac{-120}{40}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+65}{2*20}=\frac{10}{40}
=1/4
$