Answer :
[tex]2y^2-10y + 8 = 0\\ \\a=2, \ b=-10 , \ c= 8 \\ \\\Delta =b^2-4ac = (-10)^2 -4\cdot2\cdot 8 = 100-64=36 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{10-\sqrt{36}}{2\cdot 2 }=\frac{ 10-6}{4}=\frac{4}{4}= 1\\ \\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{10+\sqrt{36}}{2\cdot 2 }=\frac{ 10+6}{4}=\frac{16}{4}= 4[/tex]
