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

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

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

We multiply parentheses ..

(+2r^2-10r-5r+25)=0

We get rid of parentheses

2r^2-10r-5r+25=0

We add all the numbers together, and all the variables

2r^2-15r+25=0

a = 2; b = -15; c = +25;
Δ = b2-4ac
Δ = -152-4·2·25
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5}{2*2}=\frac{10}{4}
=2+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5}{2*2}=\frac{20}{4}
=5
$