(w-8)(w+5)=14+w+w

Solution for (w-8)(w+5)=14+w+w equation:

(w-8)(w+5)=14+w+w

We move all terms to the left:

(w-8)(w+5)-(14+w+w)=0

We add all the numbers together, and all the variables

(w-8)(w+5)-(2w+14)=0

We get rid of parentheses

(w-8)(w+5)-2w-14=0

We multiply parentheses ..

(+w^2+5w-8w-40)-2w-14=0

We get rid of parentheses

w^2+5w-8w-2w-40-14=0

We add all the numbers together, and all the variables

w^2-5w-54=0

a = 1; b = -5; c = -54;
Δ = b2-4ac
Δ = -52-4·1·(-54)
Δ = 241
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{241}}{2*1}=\frac{5-\sqrt{241}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{241}}{2*1}=\frac{5+\sqrt{241}}{2}
$