2w^2+1w-98=0

Solution for 2w^2+1w-98=0 equation:

2w^2+1w-98=0

We add all the numbers together, and all the variables

2w^2+w-98=0

a = 2; b = 1; c = -98;
Δ = b2-4ac
Δ = 12-4·2·(-98)
Δ = 785
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{785}}{2*2}=\frac{-1-\sqrt{785}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{785}}{2*2}=\frac{-1+\sqrt{785}}{4}
$