2p^2-16p+1=0

Solution for 2p^2-16p+1=0 equation:

2p^2-16p+1=0

a = 2; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·2·1
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{62}}{2*2}=\frac{16-2\sqrt{62}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{62}}{2*2}=\frac{16+2\sqrt{62}}{4}
$