N(p)=100-3p^2

Solution for N(p)=100-3p^2 equation:

(N)=100-3N^2

We move all terms to the left:

(N)-(100-3N^2)=0

We get rid of parentheses

3N^2+N-100=0

a = 3; b = 1; c = -100;
Δ = b2-4ac
Δ = 12-4·3·(-100)
Δ = 1201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1201}}{2*3}=\frac{-1-\sqrt{1201}}{6}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1201}}{2*3}=\frac{-1+\sqrt{1201}}{6}
$