Y-2(y-1)=-3(y-2)y=4+-1y

Solution for Y-2(y-1)=-3(y-2)y=4+-1y equation:

-2(Y-1)=-3(Y-2)Y=4+-1Y

We move all terms to the left:

-2(Y-1)-(-3(Y-2)Y)=0

We multiply parentheses

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


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

-3(Y-2)Y

We multiply parentheses

-3Y^2+6Y

Back to the equation:

-(-3Y^2+6Y)


We get rid of parentheses

3Y^2-6Y-2Y+2=0

We add all the numbers together, and all the variables

3Y^2-8Y+2=0

a = 3; b = -8; c = +2;
Δ = b2-4ac
Δ = -82-4·3·2
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{10}}{2*3}=\frac{8-2\sqrt{10}}{6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{10}}{2*3}=\frac{8+2\sqrt{10}}{6}
$