3(v^2+1)-v=2v+9

Solution for 3(v^2+1)-v=2v+9 equation:

3(v^2+1)-v=2v+9

We move all terms to the left:

3(v^2+1)-v-(2v+9)=0

We add all the numbers together, and all the variables

-1v+3(v^2+1)-(2v+9)=0

We multiply parentheses

3v^2-1v-(2v+9)+3=0

We get rid of parentheses

3v^2-1v-2v-9+3=0

We add all the numbers together, and all the variables

3v^2-3v-6=0

a = 3; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·3·(-6)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*3}=\frac{-6}{6}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*3}=\frac{12}{6}
=2
$