(5r-6)(2r-5)=0

Solution for (5r-6)(2r-5)=0 equation:

(5r-6)(2r-5)=0

We multiply parentheses ..

(+10r^2-25r-12r+30)=0

We get rid of parentheses

10r^2-25r-12r+30=0

We add all the numbers together, and all the variables

10r^2-37r+30=0

a = 10; b = -37; c = +30;
Δ = b2-4ac
Δ = -372-4·10·30
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-37)-13}{2*10}=\frac{24}{20}
=1+1/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-37)+13}{2*10}=\frac{50}{20}
=2+1/2
$