=2y(3y-4)

Solution for =2y(3y-4) equation:

=2y(3y-4)

We move all terms to the left:

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


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

2y(3y-4)

We multiply parentheses

6y^2-8y

Back to the equation:

-(6y^2-8y)


We get rid of parentheses

-6y^2+8y=0

a = -6; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-6)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-6}=\frac{-16}{-12}
=1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-6}=\frac{0}{-12}
=0
$