16-p^2=2p^2-4p

Solution for 16-p^2=2p^2-4p equation:

16-p^2=2p^2-4p

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1p^2-(2p^2-4p)+16=0

We get rid of parentheses

-1p^2-2p^2+4p+16=0

We add all the numbers together, and all the variables

-3p^2+4p+16=0

a = -3; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·(-3)·16
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{13}}{2*-3}=\frac{-4-4\sqrt{13}}{-6}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{13}}{2*-3}=\frac{-4+4\sqrt{13}}{-6}
$