2w^2+6w-520=0

Solution for 2w^2+6w-520=0 equation:

2w^2+6w-520=0

a = 2; b = 6; c = -520;
Δ = b2-4ac
Δ = 62-4·2·(-520)
Δ = 4196
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:

$\sqrt{\Delta}=\sqrt{4196}=\sqrt{4*1049}=\sqrt{4}*\sqrt{1049}=2\sqrt{1049}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{1049}}{2*2}=\frac{-6-2\sqrt{1049}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{1049}}{2*2}=\frac{-6+2\sqrt{1049}}{4}
$