n^2=10n+16

Solution for n^2=10n+16 equation:

n^2=10n+16

We move all terms to the left:

n^2-(10n+16)=0

We get rid of parentheses

n^2-10n-16=0

a = 1; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·1·(-16)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*1}=\frac{10-2\sqrt{41}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*1}=\frac{10+2\sqrt{41}}{2}
$