m^2-20m+25=0

Solution for m^2-20m+25=0 equation:

m^2-20m+25=0

a = 1; b = -20; c = +25;
Δ = b2-4ac
Δ = -202-4·1·25
Δ = 300
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{3}}{2*1}=\frac{20-10\sqrt{3}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{3}}{2*1}=\frac{20+10\sqrt{3}}{2}
$