What is the square of the two differences of equation 3yy-2y-2 = 0?

What is the square of the two differences of equation 3yy-2y-2 = 0?

Let 3Y ^ 2-2y-2 = 0
According to Weida's theorem, X1 + x2 = 2 / 3, X1 * x2 = - 2 / 3
（X1+X2)^2-4X1*X2=(X1-X2)^2=28/9
Or find his root directly: X1 = (1 + √ 7) / 3, X2 = (1 - √ 7) / 3
We get (x1-x2) ^ 2 = 28 / 9

(y-0.3)/0.2=(0.7y+0.5)/0.5-1.5

1/0.2=5
1/0.5=2
therefore
5(y-0.3)=2(0.7y+0.5)-1.5
5y-1.5=1.4y+1-1.5
5y-1.4y=1+1.5-1.5
3.6y=1
y=1÷3.6
y=5/18

Calculation (y-0.3) / 2 = (0.7y + 0.5) / 0.5-1.5
It should be (y-0.3) / 0.2 = (0.7y + 0.5) / 0.5-1.5

(y-0.3)/2=(0.7y+0.5)/0.5-1.5
(y-0.3)/2=2(0.7y+0.5)-1.5
Double two on both sides
y-0.3=2.8y+2-6
2.8y-y=6-2-0.3
1.8y=3.7
y=37/18