n2-10=38

Solution for n2-10=38 equation:

n2-10=38

We move all terms to the left:

n2-10-(38)=0

We add all the numbers together, and all the variables

n^2-48=0

a = 1; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·1·(-48)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*1}=\frac{0-8\sqrt{3}}{2}
=-\frac{8\sqrt{3}}{2}
=-4\sqrt{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*1}=\frac{0+8\sqrt{3}}{2}
=\frac{8\sqrt{3}}{2}
=4\sqrt{3}
$