2w^2+10w-651=0

Solution for 2w^2+10w-651=0 equation:

2w^2+10w-651=0

a = 2; b = 10; c = -651;
Δ = b2-4ac
Δ = 102-4·2·(-651)
Δ = 5308
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:

$\sqrt{\Delta}=\sqrt{5308}=\sqrt{4*1327}=\sqrt{4}*\sqrt{1327}=2\sqrt{1327}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1327}}{2*2}=\frac{-10-2\sqrt{1327}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1327}}{2*2}=\frac{-10+2\sqrt{1327}}{4}
$