n^2+101n=5400

Solution for n^2+101n=5400 equation:

n^2+101n=5400

We move all terms to the left:

n^2+101n-(5400)=0

a = 1; b = 101; c = -5400;
Δ = b2-4ac
Δ = 1012-4·1·(-5400)
Δ = 31801
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:

$\sqrt{\Delta}=\sqrt{31801}=\sqrt{49*649}=\sqrt{49}*\sqrt{649}=7\sqrt{649}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(101)-7\sqrt{649}}{2*1}=\frac{-101-7\sqrt{649}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(101)+7\sqrt{649}}{2*1}=\frac{-101+7\sqrt{649}}{2}
$