-4v(13+4v)=-3(5-9v)+2

Solution for -4v(13+4v)=-3(5-9v)+2 equation:

-4v(13+4v)=-3(5-9v)+2

We move all terms to the left:

-4v(13+4v)-(-3(5-9v)+2)=0

We add all the numbers together, and all the variables

-4v(4v+13)-(-3(-9v+5)+2)=0

We multiply parentheses

-16v^2-52v-(-3(-9v+5)+2)=0


We calculate terms in parentheses: -(-3(-9v+5)+2), so:

-3(-9v+5)+2

We multiply parentheses

27v-15+2

We add all the numbers together, and all the variables

27v-13

Back to the equation:

-(27v-13)


We get rid of parentheses

-16v^2-52v-27v+13=0

We add all the numbers together, and all the variables

-16v^2-79v+13=0

a = -16; b = -79; c = +13;
Δ = b2-4ac
Δ = -792-4·(-16)·13
Δ = 7073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-79)-\sqrt{7073}}{2*-16}=\frac{79-\sqrt{7073}}{-32}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-79)+\sqrt{7073}}{2*-16}=\frac{79+\sqrt{7073}}{-32}
$