2p^2-20p+25=0

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

2p^2-20p+25=0

a = 2; b = -20; c = +25;
Δ = b2-4ac
Δ = -202-4·2·25
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{2}}{2*2}=\frac{20-10\sqrt{2}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{2}}{2*2}=\frac{20+10\sqrt{2}}{4}
$