10r^2-25r+10=0

Solution for 10r^2-25r+10=0 equation:

10r^2-25r+10=0

a = 10; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·10·10
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15}{2*10}=\frac{10}{20}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15}{2*10}=\frac{40}{20}
=2
$