(y+3)(1-3y)=-1+2y^2;(2+根号3)y^2=(2-根号3)y;(3x+2)^2-2x^2=0因式分解解

(y+3)(1-3y)=-1+2y²
3-7y-9y²+1-2y²=0
-11y²-7y+4=0
11y²+7y-4=0
(11y-4)(y+1)=0
11y-4=0 y+1=0
∴y1=4/11
y2=-1
(2+√3)y²=(2-√3)y
2y²+√3y²-2y+√3y=0
y(2y+√3y-2+√3)=0
y=0 2y+√3y=2-√3
∴y1=0
y2=7-4√3
(3x+2)²-2x²=0
(3x+2-2x)(3x+2+2x)=0
x+2=0 5x+2=0
∴x1=-2
x2=-2/5