2w^2+28=15w

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

2w^2+28=15w

We move all terms to the left:

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

a = 2; b = -15; c = +28;
Δ = b2-4ac
Δ = -152-4·2·28
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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