3(v+1)+11=-2v(v+13)

Solution for 3(v+1)+11=-2v(v+13) equation:

3(v+1)+11=-2v(v+13)

We move all terms to the left:

3(v+1)+11-(-2v(v+13))=0

We multiply parentheses

3v-(-2v(v+13))+3+11=0


We calculate terms in parentheses: -(-2v(v+13)), so:

-2v(v+13)

We multiply parentheses

-2v^2-26v

Back to the equation:

-(-2v^2-26v)


We add all the numbers together, and all the variables

-(-2v^2-26v)+3v+14=0

We get rid of parentheses

2v^2+26v+3v+14=0

We add all the numbers together, and all the variables

2v^2+29v+14=0

a = 2; b = 29; c = +14;
Δ = b2-4ac
Δ = 292-4·2·14
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*2}=\frac{-56}{4}
=-14
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*2}=\frac{-2}{4}
=-1/2
$