(W-3)(w-3)=2w^2-3w-1

Solution for (W-3)(w-3)=2w^2-3w-1 equation:

(-3)(W-3)=2W^2-3W-1

We move all terms to the left:

(-3)(W-3)-(2W^2-3W-1)=0

We get rid of parentheses

-2W^2+(-3)(W-3)+3W+1=0

We multiply parentheses ..

-2W^2+(-3W+9)+3W+1=0

We add all the numbers together, and all the variables

-2W^2+3W+(-3W+9)+1=0

We get rid of parentheses

-2W^2+3W-3W+9+1=0

We add all the numbers together, and all the variables

-2W^2+10=0

a = -2; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-2)·10
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-2}=\frac{0-4\sqrt{5}}{-4}
=-\frac{4\sqrt{5}}{-4}
=-\frac{\sqrt{5}}{-1}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-2}=\frac{0+4\sqrt{5}}{-4}
=\frac{4\sqrt{5}}{-4}
=\frac{\sqrt{5}}{-1}
$