Nx2+3=13

Solution for Nx2+3=13 equation:

N2+3=13

We move all terms to the left:

N2+3-(13)=0

We add all the numbers together, and all the variables

N^2-10=0

a = 1; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·1·(-10)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*1}=\frac{0-2\sqrt{10}}{2}
=-\frac{2\sqrt{10}}{2}
=-\sqrt{10}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*1}=\frac{0+2\sqrt{10}}{2}
=\frac{2\sqrt{10}}{2}
=\sqrt{10}
$