n^2+18n-47=0

Solution for n^2+18n-47=0 equation:

n^2+18n-47=0

a = 1; b = 18; c = -47;
Δ = b2-4ac
Δ = 182-4·1·(-47)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-16\sqrt{2}}{2*1}=\frac{-18-16\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+16\sqrt{2}}{2*1}=\frac{-18+16\sqrt{2}}{2}
$