n^2+18n-68=0

Solution for n^2+18n-68=0 equation:

n^2+18n-68=0

a = 1; b = 18; c = -68;
Δ = b2-4ac
Δ = 182-4·1·(-68)
Δ = 596
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:

$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{149}}{2*1}=\frac{-18-2\sqrt{149}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{149}}{2*1}=\frac{-18+2\sqrt{149}}{2}
$