3(3v+1)4v+2=5(v-5)+3v

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



3(3v+1)4v+2=5(v-5)+3v

We move all terms to the left:

3(3v+1)4v+2-(5(v-5)+3v)=0

We multiply parentheses

36v^2+12v-(5(v-5)+3v)+2=0



We calculate terms in parentheses: -(5(v-5)+3v), so:

5(v-5)+3v

We add all the numbers together, and all the variables

3v+5(v-5)

We multiply parentheses

3v+5v-25

We add all the numbers together, and all the variables

8v-25

Back to the equation:

-(8v-25)



We get rid of parentheses

36v^2+12v-8v+25+2=0

We add all the numbers together, and all the variables

36v^2+4v+27=0

a = 36; b = 4; c = +27;
Δ = b2-4ac
Δ = 42-4·36·27
Δ = -3872
Delta is less than zero, so there is no solution for the equation

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