(5r-8)(4r-6)=0

Solution for (5r-8)(4r-6)=0 equation:

(5r-8)(4r-6)=0

We multiply parentheses ..

(+20r^2-30r-32r+48)=0

We get rid of parentheses

20r^2-30r-32r+48=0

We add all the numbers together, and all the variables

20r^2-62r+48=0

a = 20; b = -62; c = +48;
Δ = b2-4ac
Δ = -622-4·20·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-2}{2*20}=\frac{60}{40}
=1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+2}{2*20}=\frac{64}{40}
=1+3/5
$