5n^2+300n-5550=0

Solution for 5n^2+300n-5550=0 equation:

5n^2+300n-5550=0

a = 5; b = 300; c = -5550;
Δ = b2-4ac
Δ = 3002-4·5·(-5550)
Δ = 201000
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:

$\sqrt{\Delta}=\sqrt{201000}=\sqrt{100*2010}=\sqrt{100}*\sqrt{2010}=10\sqrt{2010}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-10\sqrt{2010}}{2*5}=\frac{-300-10\sqrt{2010}}{10}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+10\sqrt{2010}}{2*5}=\frac{-300+10\sqrt{2010}}{10}
$