Y=(3x^2+2x)

Solution for Y=(3x^2+2x) equation:

=(3Y^2+2Y)

We move all terms to the left:

-((3Y^2+2Y))=0


We calculate terms in parentheses: -((3Y^2+2Y)), so:

(3Y^2+2Y)

We get rid of parentheses

3Y^2+2Y

Back to the equation:

-(3Y^2+2Y)


We get rid of parentheses

-3Y^2-2Y=0

a = -3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-3}=\frac{0}{-6}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-3}=\frac{4}{-6}
=-2/3
$