n^2+21n-6930=0

Solution for n^2+21n-6930=0 equation:

n^2+21n-6930=0

a = 1; b = 21; c = -6930;
Δ = b2-4ac
Δ = 212-4·1·(-6930)
Δ = 28161
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:

$\sqrt{\Delta}=\sqrt{28161}=\sqrt{9*3129}=\sqrt{9}*\sqrt{3129}=3\sqrt{3129}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{3129}}{2*1}=\frac{-21-3\sqrt{3129}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{3129}}{2*1}=\frac{-21+3\sqrt{3129}}{2}
$