p^2-16p+64=3

Solution for p^2-16p+64=3 equation:

p^2-16p+64=3

We move all terms to the left:

p^2-16p+64-(3)=0

We add all the numbers together, and all the variables

p^2-16p+61=0

a = 1; b = -16; c = +61;
Δ = b2-4ac
Δ = -162-4·1·61
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{3}}{2*1}=\frac{16-2\sqrt{3}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{3}}{2*1}=\frac{16+2\sqrt{3}}{2}
$