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

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

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

We multiply parentheses ..

(+5r^2-10r-2r+4)=0

We get rid of parentheses

5r^2-10r-2r+4=0

We add all the numbers together, and all the variables

5r^2-12r+4=0

a = 5; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·5·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*5}=\frac{4}{10}
=2/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*5}=\frac{20}{10}
=2
$