10(v+4)2v=2(v+1)-8

Solution for 10(v+4)2v=2(v+1)-8 equation:

10(v+4)2v=2(v+1)-8

We move all terms to the left:

10(v+4)2v-(2(v+1)-8)=0

We multiply parentheses

20v^2+80v-(2(v+1)-8)=0


We calculate terms in parentheses: -(2(v+1)-8), so:

2(v+1)-8

We multiply parentheses

2v+2-8

We add all the numbers together, and all the variables

2v-6

Back to the equation:

-(2v-6)


We get rid of parentheses

20v^2+80v-2v+6=0

We add all the numbers together, and all the variables

20v^2+78v+6=0

a = 20; b = 78; c = +6;
Δ = b2-4ac
Δ = 782-4·20·6
Δ = 5604
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:

$\sqrt{\Delta}=\sqrt{5604}=\sqrt{4*1401}=\sqrt{4}*\sqrt{1401}=2\sqrt{1401}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(78)-2\sqrt{1401}}{2*20}=\frac{-78-2\sqrt{1401}}{40}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(78)+2\sqrt{1401}}{2*20}=\frac{-78+2\sqrt{1401}}{40}
$