(W+3)(w+3)+4=(w-2)(w-2)+5w-2

Solution for (W+3)(w+3)+4=(w-2)(w-2)+5w-2 equation:

(+3)(W+3)+4=(W-2)(W-2)+5W-2

We move all terms to the left:

(+3)(W+3)+4-((W-2)(W-2)+5W-2)=0

We add all the numbers together, and all the variables

3(W+3)-((W-2)(W-2)+5W-2)+4=0

We multiply parentheses

3W-((W-2)(W-2)+5W-2)+9+4=0

We multiply parentheses ..

-((+W^2-2W-2W+4)+5W-2)+3W+9+4=0


We calculate terms in parentheses: -((+W^2-2W-2W+4)+5W-2), so:

(+W^2-2W-2W+4)+5W-2

We get rid of parentheses

W^2-2W-2W+5W+4-2

We add all the numbers together, and all the variables

W^2+W+2

Back to the equation:

-(W^2+W+2)


We add all the numbers together, and all the variables

3W-(W^2+W+2)+13=0

We get rid of parentheses

-W^2+3W-W-2+13=0

We add all the numbers together, and all the variables

-1W^2+2W+11=0

a = -1; b = 2; c = +11;
Δ = b2-4ac
Δ = 22-4·(-1)·11
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{3}}{2*-1}=\frac{-2-4\sqrt{3}}{-2}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{3}}{2*-1}=\frac{-2+4\sqrt{3}}{-2}
$