(17v^2)+35v=-8+(8v^2)+8v

Solution for (17v^2)+35v=-8+(8v^2)+8v equation:

(17v^2)+35v=-8+(8v^2)+8v

We move all terms to the left:

(17v^2)+35v-(-8+(8v^2)+8v)=0

determiningTheFunctionDomain
17v^2-(-8+8v^2+8v)+35v=0

We get rid of parentheses

17v^2-8v^2-8v+35v+8=0

We add all the numbers together, and all the variables

9v^2+27v+8=0

a = 9; b = 27; c = +8;
Δ = b2-4ac
Δ = 272-4·9·8
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-21}{2*9}=\frac{-48}{18}
=-2+2/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+21}{2*9}=\frac{-6}{18}
=-1/3
$