(w+2)(w)=28

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

(w+2)(w)=28

We move all terms to the left:

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

We multiply parentheses

w^2+2w-28=0

a = 1; b = 2; c = -28;
Δ = b2-4ac
Δ = 22-4·1·(-28)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{29}}{2*1}=\frac{-2-2\sqrt{29}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{29}}{2*1}=\frac{-2+2\sqrt{29}}{2}
$