(2w+1)(w)=28^2

Solution for (2w+1)(w)=28^2 equation:

(2w+1)(w)=28^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

(2w+1)w-784=0

We multiply parentheses

2w^2+w-784=0

a = 2; b = 1; c = -784;
Δ = b2-4ac
Δ = 12-4·2·(-784)
Δ = 6273
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:

$\sqrt{\Delta}=\sqrt{6273}=\sqrt{9*697}=\sqrt{9}*\sqrt{697}=3\sqrt{697}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{697}}{2*2}=\frac{-1-3\sqrt{697}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{697}}{2*2}=\frac{-1+3\sqrt{697}}{4}
$