D=-3p^2+162p-255

Solution for D=-3p^2+162p-255 equation:

=-3D^2+162D-255

We move all terms to the left:

-(-3D^2+162D-255)=0

We get rid of parentheses

3D^2-162D+255=0

a = 3; b = -162; c = +255;
Δ = b2-4ac
Δ = -1622-4·3·255
Δ = 23184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{23184}=\sqrt{144*161}=\sqrt{144}*\sqrt{161}=12\sqrt{161}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-162)-12\sqrt{161}}{2*3}=\frac{162-12\sqrt{161}}{6}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-162)+12\sqrt{161}}{2*3}=\frac{162+12\sqrt{161}}{6}
$