2w^2+16w-94=0

Solution for 2w^2+16w-94=0 equation:

2w^2+16w-94=0

a = 2; b = 16; c = -94;
Δ = b2-4ac
Δ = 162-4·2·(-94)
Δ = 1008
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:

$\sqrt{\Delta}=\sqrt{1008}=\sqrt{144*7}=\sqrt{144}*\sqrt{7}=12\sqrt{7}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-12\sqrt{7}}{2*2}=\frac{-16-12\sqrt{7}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+12\sqrt{7}}{2*2}=\frac{-16+12\sqrt{7}}{4}
$