(W+2)(w+10)=(w^2+12)

Solution for (W+2)(w+10)=(w^2+12) equation:

(+2)(W+10)=(W^2+12)

We move all terms to the left:

(+2)(W+10)-((W^2+12))=0

We add all the numbers together, and all the variables

2(W+10)-((W^2+12))=0

We multiply parentheses

2W-((W^2+12))+20=0


We calculate terms in parentheses: -((W^2+12)), so:

(W^2+12)

We get rid of parentheses

W^2+12

Back to the equation:

-(W^2+12)


We get rid of parentheses

-W^2+2W-12+20=0

We add all the numbers together, and all the variables

-1W^2+2W+8=0

a = -1; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-1}=\frac{-8}{-2}
=+4
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-1}=\frac{4}{-2}
=-2
$