n^2+16n-18=0

Solution for n^2+16n-18=0 equation:

n^2+16n-18=0

a = 1; b = 16; c = -18;
Δ = b2-4ac
Δ = 162-4·1·(-18)
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{82}}{2*1}=\frac{-16-2\sqrt{82}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{82}}{2*1}=\frac{-16+2\sqrt{82}}{2}
$