0=(Y^2-4y+9)-(3y^2-6y-9)

Solution for 0=(Y^2-4y+9)-(3y^2-6y-9) equation:

0=(^2-4Y+9)-(3Y^2-6Y-9)

We move all terms to the left:

0-((^2-4Y+9)-(3Y^2-6Y-9))=0

We add all the numbers together, and all the variables

-((^2-4Y+9)-(3Y^2-6Y-9))=0


We calculate terms in parentheses: -((^2-4Y+9)-(3Y^2-6Y-9)), so:

(^2-4Y+9)-(3Y^2-6Y-9)

We get rid of parentheses

-3Y^2-4Y+6Y+9+9+^2

We add all the numbers together, and all the variables

-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
$