Answer
$\displaystyle \{\frac{1}{4},9\}$
Work Step by Step
Substitute $u=\sqrt{y}, u^{2}=y,$
$2u^{2}-7u+3=0$
... ... factoring, we seek factors of $ac=6$ whose sum is $b=-7$
... we find $-6$ and $-1$
$2u^{2}-6u-u+3=0$
$2u(u-3)-(u-3)=0$
$(u-3)(2u-1)=0$
$u=3$ or $u=1/2$
... bring back $y$
$\sqrt{y}=3 \qquad$or$\qquad \displaystyle \sqrt{y}=\frac{1}{2}$
$y=3^{2} \qquad$or$\qquad y=(\displaystyle \frac{1}{2})^{2}$
$y=9 \qquad$or$\qquad y=\displaystyle \frac{1}{4}$
Solution set = $\displaystyle \{\frac{1}{4},9\}$
