N(t)=4t^2+7t-2

Solution for N(t)=4t^2+7t-2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

-4N^2+N-7N+2=0

We add all the numbers together, and all the variables

-4N^2-6N+2=0

a = -4; b = -6; c = +2;
Δ = b2-4ac
Δ = -62-4·(-4)·2
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*-4}=\frac{6-2\sqrt{17}}{-8}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*-4}=\frac{6+2\sqrt{17}}{-8}
$