(2w-3)(w-5)=55

Solution for (2w-3)(w-5)=55 equation:

(2w-3)(w-5)=55

We move all terms to the left:

(2w-3)(w-5)-(55)=0

We multiply parentheses ..

(+2w^2-10w-3w+15)-55=0

We get rid of parentheses

2w^2-10w-3w+15-55=0

We add all the numbers together, and all the variables

2w^2-13w-40=0

a = 2; b = -13; c = -40;
Δ = b2-4ac
Δ = -132-4·2·(-40)
Δ = 489
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{489}}{2*2}=\frac{13-\sqrt{489}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{489}}{2*2}=\frac{13+\sqrt{489}}{4}
$