2w^2+18w-129=0

Solution for 2w^2+18w-129=0 equation:

2w^2+18w-129=0

a = 2; b = 18; c = -129;
Δ = b2-4ac
Δ = 182-4·2·(-129)
Δ = 1356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1356}=\sqrt{4*339}=\sqrt{4}*\sqrt{339}=2\sqrt{339}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{339}}{2*2}=\frac{-18-2\sqrt{339}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{339}}{2*2}=\frac{-18+2\sqrt{339}}{4}
$