x+x+2+x+4=36-x(x+2)

Solution for x+x+2+x+4=36-x(x+2) equation:

x+x+2+x+4=36-x(x+2)

We move all terms to the left:

x+x+2+x+4-(36-x(x+2))=0

We add all the numbers together, and all the variables

3x-(36-x(x+2))+6=0


We calculate terms in parentheses: -(36-x(x+2)), so:

36-x(x+2)

determiningTheFunctionDomain
-x(x+2)+36

We multiply parentheses

-x^2-2x+36

We add all the numbers together, and all the variables

-1x^2-2x+36

Back to the equation:

-(-1x^2-2x+36)


We get rid of parentheses

1x^2+2x+3x-36+6=0

We add all the numbers together, and all the variables

x^2+5x-30=0

a = 1; b = 5; c = -30;
Δ = b2-4ac
Δ = 52-4·1·(-30)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{145}}{2*1}=\frac{-5-\sqrt{145}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{145}}{2*1}=\frac{-5+\sqrt{145}}{2}
$