2w^2+w=68

Solution for 2w^2+w=68 equation:

2w^2+w=68

We move all terms to the left:

2w^2+w-(68)=0

a = 2; b = 1; c = -68;
Δ = b2-4ac
Δ = 12-4·2·(-68)
Δ = 545
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{545}}{2*2}=\frac{-1-\sqrt{545}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{545}}{2*2}=\frac{-1+\sqrt{545}}{4}
$