n^2-3n-405=0

Solution for n^2-3n-405=0 equation:

n^2-3n-405=0

a = 1; b = -3; c = -405;
Δ = b2-4ac
Δ = -32-4·1·(-405)
Δ = 1629
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:

$\sqrt{\Delta}=\sqrt{1629}=\sqrt{9*181}=\sqrt{9}*\sqrt{181}=3\sqrt{181}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{181}}{2*1}=\frac{3-3\sqrt{181}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{181}}{2*1}=\frac{3+3\sqrt{181}}{2}
$