(5r-9)(4r-6)=0

Solution for (5r-9)(4r-6)=0 equation:

(5r-9)(4r-6)=0

We multiply parentheses ..

(+20r^2-30r-36r+54)=0

We get rid of parentheses

20r^2-30r-36r+54=0

We add all the numbers together, and all the variables

20r^2-66r+54=0

a = 20; b = -66; c = +54;
Δ = b2-4ac
Δ = -662-4·20·54
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-66)-6}{2*20}=\frac{60}{40}
=1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-66)+6}{2*20}=\frac{72}{40}
=1+4/5
$