Question

14y - 5y^2 +24 =0

Answer 1

\(\Leftrightarrow\left(x-4\right)\left(x+\dfrac{6}{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-6}{5}\end{matrix}\right.\)

Vậy pt có nghiệm là S=\(\left\{4;\dfrac{-6}{5}\right\}\)

Answer 2

=>5y^2-14y-24=0

=>5y^2-20y+6y-24=0

=>(y-4)(5y+6)=0

=>y=-6/5 hoặc y=4