N(t)=-4t2+40t+25

Solution for N(t)=-4t2+40t+25 equation:

(N)=-4N^2+40N+25

We move all terms to the left:

(N)-(-4N^2+40N+25)=0

We get rid of parentheses

4N^2-40N+N-25=0

We add all the numbers together, and all the variables

4N^2-39N-25=0

a = 4; b = -39; c = -25;
Δ = b2-4ac
Δ = -392-4·4·(-25)
Δ = 1921
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{-(-39)-\sqrt{1921}}{2*4}=\frac{39-\sqrt{1921}}{8}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1921}}{2*4}=\frac{39+\sqrt{1921}}{8}
$