6(-3v+1)=(-2v2)

Solution for 6(-3v+1)=(-2v2) equation:

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

We move all terms to the left:

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

We multiply parentheses

-((-2v^2))-18v+6=0


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

(-2v^2)

We get rid of parentheses

-2v^2

Back to the equation:

-(-2v^2)


We get rid of parentheses

2v^2-18v+6=0

a = 2; b = -18; c = +6;
Δ = b2-4ac
Δ = -182-4·2·6
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{69}}{2*2}=\frac{18-2\sqrt{69}}{4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{69}}{2*2}=\frac{18+2\sqrt{69}}{4}
$