130=3y(4y+2)

Solution for 130=3y(4y+2) equation:

130=3y(4y+2)

We move all terms to the left:

130-(3y(4y+2))=0


We calculate terms in parentheses: -(3y(4y+2)), so:

3y(4y+2)

We multiply parentheses

12y^2+6y

Back to the equation:

-(12y^2+6y)


We get rid of parentheses

-12y^2-6y+130=0

a = -12; b = -6; c = +130;
Δ = b2-4ac
Δ = -62-4·(-12)·130
Δ = 6276
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6276}=\sqrt{4*1569}=\sqrt{4}*\sqrt{1569}=2\sqrt{1569}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{1569}}{2*-12}=\frac{6-2\sqrt{1569}}{-24}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{1569}}{2*-12}=\frac{6+2\sqrt{1569}}{-24}
$