2w^2+6w-200=0

Solution for 2w^2+6w-200=0 equation:

2w^2+6w-200=0

a = 2; b = 6; c = -200;
Δ = b2-4ac
Δ = 62-4·2·(-200)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{409}}{2*2}=\frac{-6-2\sqrt{409}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{409}}{2*2}=\frac{-6+2\sqrt{409}}{4}
$