P^2-16p+13=-3p^2

Solution for P^2-16p+13=-3p^2 equation:

^2-16P+13=-3P^2

We move all terms to the left:

^2-16P+13-(-3P^2)=0

We add all the numbers together, and all the variables

-(-3P^2)-16P=0

We get rid of parentheses

3P^2-16P=0

a = 3; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·3·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*3}=\frac{0}{6}
=0
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*3}=\frac{32}{6}
=5+1/3
$