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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

w^2-13w+26=0

a = 1; b = -13; c = +26;
Δ = b2-4ac
Δ = -132-4·1·26
Δ = 65
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{-(-13)-\sqrt{65}}{2*1}=\frac{13-\sqrt{65}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{65}}{2*1}=\frac{13+\sqrt{65}}{2}
$