4(v+1)1v=3(v-2)+2

Solution for 4(v+1)1v=3(v-2)+2 equation:

4(v+1)1v=3(v-2)+2

We move all terms to the left:

4(v+1)1v-(3(v-2)+2)=0

We multiply parentheses

4v^2+4v-(3(v-2)+2)=0


We calculate terms in parentheses: -(3(v-2)+2), so:

3(v-2)+2

We multiply parentheses

3v-6+2

We add all the numbers together, and all the variables

3v-4

Back to the equation:

-(3v-4)


We get rid of parentheses

4v^2+4v-3v+4=0

We add all the numbers together, and all the variables

4v^2+v+4=0

a = 4; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·4·4
Δ = -63
Delta is less than zero, so there is no solution for the equation