2w^2+6w=28

Solution for 2w^2+6w=28 equation:

2w^2+6w=28

We move all terms to the left:

2w^2+6w-(28)=0

a = 2; b = 6; c = -28;
Δ = b2-4ac
Δ = 62-4·2·(-28)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{65}}{2*2}=\frac{-6-2\sqrt{65}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{65}}{2*2}=\frac{-6+2\sqrt{65}}{4}
$