(3v-2v-5)=(4v2-7v+1)

Solution for (3v-2v-5)=(4v2-7v+1) equation:

(3v-2v-5)=(4v^2-7v+1)

We move all terms to the left:

(3v-2v-5)-((4v^2-7v+1))=0

We add all the numbers together, and all the variables

(v-5)-((4v^2-7v+1))=0

We get rid of parentheses

v-((4v^2-7v+1))-5=0


We calculate terms in parentheses: -((4v^2-7v+1)), so:

(4v^2-7v+1)

We get rid of parentheses

4v^2-7v+1

Back to the equation:

-(4v^2-7v+1)


We get rid of parentheses

-4v^2+v+7v-1-5=0

We add all the numbers together, and all the variables

-4v^2+8v-6=0

a = -4; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·(-4)·(-6)
Δ = -32
Delta is less than zero, so there is no solution for the equation