6(-3v(2)+1)=5(-2(2)-2

Solution for 6(-3v(2)+1)=5(-2(2)-2 equation:

6(-3v(2)+1)=5(-2(2)-2

We move all terms to the left:

6(-3v(2)+1)-(5(-2(2)-2)=0

We add all the numbers together, and all the variables

6(-3v^2+1)-(5(-24)=0

We add all the numbers together, and all the variables

6(-3v^2+1)=0

We multiply parentheses

-18v^2+6=0

a = -18; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-18)·6
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*-18}=\frac{0-12\sqrt{3}}{-36}
=-\frac{12\sqrt{3}}{-36}
=-\frac{\sqrt{3}}{-3}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-18}=\frac{0+12\sqrt{3}}{-36}
=\frac{12\sqrt{3}}{-36}
=\frac{\sqrt{3}}{-3}
$