3y-3/3y-2+6+2y/3y+2=2

Решите задачу:

$\frac{3y-3}{3y-2}+ \frac{6+2y}{3y+2}=2$
$\frac{(3y+2)*(3y-3)+(3y-2)*(6+2y)}{(3y-2)*(3y+2)}=2$
$\frac{9y ^{2}-9y+6y-6+18y+6y ^{2}-12-4y }{9y ^{2}-4 }=2$
$\frac{15y ^{2}+11y-18 }{9y ^{2}-4 }=2$
$15y ^{2}+11y-18=2*(9y ^{2}-4)$
$15y ^{2}+11y-18=18y ^{2}-8$
$15y ^{2}+11y-18-18y ^{2}+8=0$
$-3y ^{2}+11y-10=0$
$D=b ^{2}-4ac=11 ^{2}-4*(-3)*(-10)=121-120=1$
$y _{1}= \frac{-b+ \sqrt{D} }{2a}= \frac{-11+1}{2*(-3)}= \frac{-10}{-6}= \frac{5}{3}=1 \frac{2}{3}$ $y _{2}= \frac{-b- \sqrt{D} }{2a}= \frac{-11-1}{-6}= \frac{-12}{-6}=2$