(w-5)(w-5)=2w^2-18w+40

Solution for (w-5)(w-5)=2w^2-18w+40 equation:

(w-5)(w-5)=2w^2-18w+40

We move all terms to the left:

(w-5)(w-5)-(2w^2-18w+40)=0

We get rid of parentheses

-2w^2+(w-5)(w-5)+18w-40=0

We multiply parentheses ..

-2w^2+(+w^2-5w-5w+25)+18w-40=0

We get rid of parentheses

-2w^2+w^2-5w-5w+18w+25-40=0

We add all the numbers together, and all the variables

-1w^2+8w-15=0

a = -1; b = 8; c = -15;
Δ = b2-4ac
Δ = 82-4·(-1)·(-15)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*-1}=\frac{-10}{-2}
=+5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*-1}=\frac{-6}{-2}
=+3
$