W(w+5)+(w+2)(w+5)=28

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

(W+5)+(W+2)(W+5)=28

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

(+W^2+5W+2W+10)+W+5-28=0

We add all the numbers together, and all the variables

(+W^2+5W+2W+10)+W-23=0

We get rid of parentheses

W^2+5W+2W+W+10-23=0

We add all the numbers together, and all the variables

W^2+8W-13=0

a = 1; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·1·(-13)
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{-(8)-2\sqrt{29}}{2*1}=\frac{-8-2\sqrt{29}}{2}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{29}}{2*1}=\frac{-8+2\sqrt{29}}{2}
$