n^2-17=16n

Solution for n^2-17=16n equation:

n^2-17=16n

We move all terms to the left:

n^2-17-(16n)=0

a = 1; b = -16; c = -17;
Δ = b2-4ac
Δ = -162-4·1·(-17)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-18}{2*1}=\frac{-2}{2}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+18}{2*1}=\frac{34}{2}
=17
$