20-p^2=4p

Solution for 20-p^2=4p equation:

20-p^2=4p

We move all terms to the left:

20-p^2-(4p)=0

We add all the numbers together, and all the variables

-1p^2-4p+20=0

a = -1; b = -4; c = +20;
Δ = b2-4ac
Δ = -42-4·(-1)·20
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{6}}{2*-1}=\frac{4-4\sqrt{6}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{6}}{2*-1}=\frac{4+4\sqrt{6}}{-2}
$