4(2v^2+17)-9v=7v(v+1)+4

Solution for 4(2v^2+17)-9v=7v(v+1)+4 equation:

4(2v^2+17)-9v=7v(v+1)+4

We move all terms to the left:

4(2v^2+17)-9v-(7v(v+1)+4)=0

We add all the numbers together, and all the variables

-9v+4(2v^2+17)-(7v(v+1)+4)=0

We multiply parentheses

8v^2-9v-(7v(v+1)+4)+68=0


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

7v(v+1)+4

We multiply parentheses

7v^2+7v+4

Back to the equation:

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


We get rid of parentheses

8v^2-7v^2-9v-7v-4+68=0

We add all the numbers together, and all the variables

v^2-16v+64=0

a = 1; b = -16; c = +64;
Δ = b2-4ac
Δ = -162-4·1·64
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$v=\frac{-b}{2a}=\frac{16}{2}=8$