11v+1-11v+7-4v=7(v+1)D4v=7(v+1)

Solution for 11v+1-11v+7-4v=7(v+1)D4v=7(v+1) equation:

11v+1-11v+7-4v=7(v+1)4v=7(v+1)

We move all terms to the left:

11v+1-11v+7-4v-(7(v+1)4v)=0

We add all the numbers together, and all the variables

-4v-(7(v+1)4v)+8=0


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

7(v+1)4v

We multiply parentheses

28v^2+28v

Back to the equation:

-(28v^2+28v)


We get rid of parentheses

-28v^2-4v-28v+8=0

We add all the numbers together, and all the variables

-28v^2-32v+8=0

a = -28; b = -32; c = +8;
Δ = b2-4ac
Δ = -322-4·(-28)·8
Δ = 1920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{30}}{2*-28}=\frac{32-8\sqrt{30}}{-56}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{30}}{2*-28}=\frac{32+8\sqrt{30}}{-56}
$