p2-12p+32=0

Solution for p2-12p+32=0 equation:

p2-12p+32=0

We add all the numbers together, and all the variables

p^2-12p+32=0

a = 1; b = -12; c = +32;
Δ = b2-4ac
Δ = -122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*1}=\frac{8}{2}
=4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*1}=\frac{16}{2}
=8
$