2w^2+15w=28

Solution for 2w^2+15w=28 equation:

2w^2+15w=28

We move all terms to the left:

2w^2+15w-(28)=0

a = 2; b = 15; c = -28;
Δ = b2-4ac
Δ = 152-4·2·(-28)
Δ = 449
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{-(15)-\sqrt{449}}{2*2}=\frac{-15-\sqrt{449}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{449}}{2*2}=\frac{-15+\sqrt{449}}{4}
$