3(u-1)=5u+4-2(-u2-1)

Solution for 3(u-1)=5u+4-2(-u2-1) equation:

3(u-1)=5u+4-2(-u2-1)

We move all terms to the left:

3(u-1)-(5u+4-2(-u2-1))=0

We add all the numbers together, and all the variables

-(5u+4-2(-1u^2-1))+3(u-1)=0

We multiply parentheses

-(5u+4-2(-1u^2-1))+3u-3=0


We calculate terms in parentheses: -(5u+4-2(-1u^2-1)), so:

5u+4-2(-1u^2-1)

determiningTheFunctionDomain
-2(-1u^2-1)+5u+4

We multiply parentheses

2u^2+5u+2+4

We add all the numbers together, and all the variables

2u^2+5u+6

Back to the equation:

-(2u^2+5u+6)


We add all the numbers together, and all the variables

3u-(2u^2+5u+6)-3=0

We get rid of parentheses

-2u^2+3u-5u-6-3=0

We add all the numbers together, and all the variables

-2u^2-2u-9=0

a = -2; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·(-2)·(-9)
Δ = -68
Delta is less than zero, so there is no solution for the equation