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

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

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

We move all terms to the left:

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


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

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

We get rid of parentheses

-3Y^2+3Y+2

Back to the equation:

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


We get rid of parentheses

3Y^2-3Y-2=0

a = 3; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·3·(-2)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*3}=\frac{3-\sqrt{33}}{6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*3}=\frac{3+\sqrt{33}}{6}
$