(5r-3)(r-8)=0

Solution for (5r-3)(r-8)=0 equation:

(5r-3)(r-8)=0

We multiply parentheses ..

(+5r^2-40r-3r+24)=0

We get rid of parentheses

5r^2-40r-3r+24=0

We add all the numbers together, and all the variables

5r^2-43r+24=0

a = 5; b = -43; c = +24;
Δ = b2-4ac
Δ = -432-4·5·24
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-37}{2*5}=\frac{6}{10}
=3/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+37}{2*5}=\frac{80}{10}
=8
$