4(3+w)-w=6+w(w+2)

Solution for 4(3+w)-w=6+w(w+2) equation:

4(3+w)-w=6+w(w+2)

We move all terms to the left:

4(3+w)-w-(6+w(w+2))=0

We add all the numbers together, and all the variables

4(w+3)-w-(6+w(w+2))=0

We add all the numbers together, and all the variables

-1w+4(w+3)-(6+w(w+2))=0

We multiply parentheses

-1w+4w-(6+w(w+2))+12=0


We calculate terms in parentheses: -(6+w(w+2)), so:

6+w(w+2)

determiningTheFunctionDomain
w(w+2)+6

We multiply parentheses

w^2+2w+6

Back to the equation:

-(w^2+2w+6)


We add all the numbers together, and all the variables

3w-(w^2+2w+6)+12=0

We get rid of parentheses

-w^2+3w-2w-6+12=0

We add all the numbers together, and all the variables

-1w^2+w+6=0

a = -1; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-1}=\frac{-6}{-2}
=+3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-1}=\frac{4}{-2}
=-2
$