Question

Решите уравнение:
(3у-2)(4у+1)-у(2-3у)=0

Answer 1

$(3y-2)(4y+1)-y(2-3y)=0\\\\(3y-2)(4y+1)+y(3y-2)=0\\\\(3y-2)(4y+1+y)=0\\\\(3y-2)(5y+1)=0\\\\\left[\begin{array}{ccc}3y-2=0\\5y+1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}3y=2\\5y=-1\end{array}\right\\\\\\\left[\begin{array}{ccc}y_{1}=\frac{2}{3}\\y_{2}=-0,2 \end{array}\right\\\\Otvet:\boxed{\frac{2}{3};-0,2}$