Question

Допоможіть будь ласка! З повним розвʼязком

Answer 1

$\displaystyle\bf\\(3y-5+2y^{2} )(\sqrt{y} -7)=0\\\\\\\left[\begin{array}{ccc}2y^{2}+3y-5=0 \\\sqrt{y} -7=0\end{array}\right\\\\\\1)\\\\2y^{2} +3y-5=0\\\\D=3^{2} -4\cdot 2 \cdot(-5)=9+40=49=7^{2} \\\\\\y_{1} =\frac{-3-7}{4} =\frac{-10}{4} =-2,5\\\\\\y_{2} =\frac{-3+7}{4} =\frac{4}{4} =1\\\\\\2)\\\\\sqrt{y} -7=0\\\\\sqrt{y}=7\\\\y_{3} =49\\\\\\Otvet \ : \ -2,5 \ ; \ 1 \ ; \ 49$