n2+(n+4)-160=8000

Solution for n2+(n+4)-160=8000 equation:

n2+(n+4)-160=8000

We move all terms to the left:

n2+(n+4)-160-(8000)=0

We add all the numbers together, and all the variables

n^2+(n+4)-8160=0

We get rid of parentheses

n^2+n+4-8160=0

We add all the numbers together, and all the variables

n^2+n-8156=0

a = 1; b = 1; c = -8156;
Δ = b2-4ac
Δ = 12-4·1·(-8156)
Δ = 32625
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:

$\sqrt{\Delta}=\sqrt{32625}=\sqrt{225*145}=\sqrt{225}*\sqrt{145}=15\sqrt{145}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15\sqrt{145}}{2*1}=\frac{-1-15\sqrt{145}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15\sqrt{145}}{2*1}=\frac{-1+15\sqrt{145}}{2}
$