N=-4p^2+20p+7

Solution for N=-4p^2+20p+7 equation:

=-4N^2+20N+7

We move all terms to the left:

-(-4N^2+20N+7)=0

We get rid of parentheses

4N^2-20N-7=0

a = 4; b = -20; c = -7;
Δ = b2-4ac
Δ = -202-4·4·(-7)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16\sqrt{2}}{2*4}=\frac{20-16\sqrt{2}}{8}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16\sqrt{2}}{2*4}=\frac{20+16\sqrt{2}}{8}
$