0=2w^2+w-28

Solution for 0=2w^2+w-28 equation:

0=2w^2+w-28

We move all terms to the left:

0-(2w^2+w-28)=0

We add all the numbers together, and all the variables

-(2w^2+w-28)=0

We get rid of parentheses

-2w^2-w+28=0

We add all the numbers together, and all the variables

-2w^2-1w+28=0

a = -2; b = -1; c = +28;
Δ = b2-4ac
Δ = -12-4·(-2)·28
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*-2}=\frac{-14}{-4}
=3+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*-2}=\frac{16}{-4}
=-4
$