41=(Nx2)+3

Solution for 41=(Nx2)+3 equation:

41=(N2)+3

We move all terms to the left:

41-((N2)+3)=0

We add all the numbers together, and all the variables

-(+N^2+3)+41=0

We get rid of parentheses

-N^2-3+41=0

We add all the numbers together, and all the variables

-1N^2+38=0

a = -1; b = 0; c = +38;
Δ = b2-4ac
Δ = 02-4·(-1)·38
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{38}}{2*-1}=\frac{0-2\sqrt{38}}{-2}
=-\frac{2\sqrt{38}}{-2}
=-\frac{\sqrt{38}}{-1}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{38}}{2*-1}=\frac{0+2\sqrt{38}}{-2}
=\frac{2\sqrt{38}}{-2}
=\frac{\sqrt{38}}{-1}
$