p^2-24p+128=0

Solution for p^2-24p+128=0 equation:

p^2-24p+128=0

a = 1; b = -24; c = +128;
Δ = b2-4ac
Δ = -242-4·1·128
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8}{2*1}=\frac{16}{2}
=8
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*1}=\frac{32}{2}
=16
$