w(2w+5)=28

Solution for w(2w+5)=28 equation:

w(2w+5)=28

We move all terms to the left:

w(2w+5)-(28)=0

We multiply parentheses

2w^2+5w-28=0

a = 2; b = 5; c = -28;
Δ = b2-4ac
Δ = 52-4·2·(-28)
Δ = 249
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{-(5)-\sqrt{249}}{2*2}=\frac{-5-\sqrt{249}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{249}}{2*2}=\frac{-5+\sqrt{249}}{4}
$