Please help me with this as well

Answer:
[tex]y = \dfrac{3 + 6\sqrt{2}}{4} ~~~\textrm{and}~~~y = \dfrac{3 - 6\sqrt{2}}{4}[/tex]
Step-by-step explanation:
[tex] (4y - 3)^2 = 72 [/tex]
[tex] 4y - 3 = \pm\sqrt{72} [/tex]
[tex] 4y - 3 = \pm 6\sqrt{2} [/tex]
[tex] 4y = 3 \pm 6\sqrt{2} [/tex]
[tex]y = \dfrac{3 + 6\sqrt{2}}{4}~~~\textrm{or}~~~y = \dfrac{3 - 6\sqrt{2}}{4}[/tex]