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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+4r^2-10r-10r+25)-5=0

We get rid of parentheses

4r^2-10r-10r+25-5=0

We add all the numbers together, and all the variables

4r^2-20r+20=0

a = 4; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·4·20
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*4}=\frac{20-4\sqrt{5}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*4}=\frac{20+4\sqrt{5}}{8}
$