8r^2-20r+8=0

Solution for 8r^2-20r+8=0 equation:

8r^2-20r+8=0

a = 8; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·8·8
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*8}=\frac{8}{16}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*8}=\frac{32}{16}
=2
$