3(2+u)-u=6+u(u+1)

Solution for 3(2+u)-u=6+u(u+1) equation:

3(2+u)-u=6+u(u+1)

We move all terms to the left:

3(2+u)-u-(6+u(u+1))=0

We add all the numbers together, and all the variables

3(u+2)-u-(6+u(u+1))=0

We add all the numbers together, and all the variables

-1u+3(u+2)-(6+u(u+1))=0

We multiply parentheses

-1u+3u-(6+u(u+1))+6=0


We calculate terms in parentheses: -(6+u(u+1)), so:

6+u(u+1)

determiningTheFunctionDomain
u(u+1)+6

We multiply parentheses

u^2+u+6

Back to the equation:

-(u^2+u+6)


We add all the numbers together, and all the variables

2u-(u^2+u+6)+6=0

We get rid of parentheses

-u^2+2u-u-6+6=0

We add all the numbers together, and all the variables

-1u^2+u=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-1}=\frac{-2}{-2}
=1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-1}=\frac{0}{-2}
=0
$