\(2y^2-5y=-2\ \ \ /+2\\\\2y^2-5y+2=0\\\\a=2;\ b=-5;\ c=2\\\\\Delta=b^2-4ac\\\\\Delta=(-5)^2-4\cdot2\cdot2=25-16=9\\\\y_1=\frac{-b-\sqrt\Delta}{2a};\ y_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt9=3\\\\y_1=\frac{5-3}{2\cdot2}=\frac{2}{4}=\frac{1}{2};\ y_2=\frac{5+3}{2\cdot2}=\frac{8}{4}=2\)