10=3+n^2

Solution for 10=3+n^2 equation:

10=3+n^2

We move all terms to the left:

10-(3+n^2)=0

We get rid of parentheses

-n^2-3+10=0

We add all the numbers together, and all the variables

-1n^2+7=0

a = -1; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-1)·7
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{7}}{2*-1}=\frac{0-2\sqrt{7}}{-2}
=-\frac{2\sqrt{7}}{-2}
=-\frac{\sqrt{7}}{-1}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{7}}{2*-1}=\frac{0+2\sqrt{7}}{-2}
=\frac{2\sqrt{7}}{-2}
=\frac{\sqrt{7}}{-1}
$