n^2+4+1=325

Solution for n^2+4+1=325 equation:

n^2+4+1=325

We move all terms to the left:

n^2+4+1-(325)=0

We add all the numbers together, and all the variables

n^2-320=0

a = 1; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·1·(-320)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*1}=\frac{0-16\sqrt{5}}{2}
=-\frac{16\sqrt{5}}{2}
=-8\sqrt{5}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*1}=\frac{0+16\sqrt{5}}{2}
=\frac{16\sqrt{5}}{2}
=8\sqrt{5}
$