how do you describe the roots of 9y² - 6y + 5 = 0?
\(9y^2- 6y + 5 = 0\\ 9y^2-6y+1+4=0\\ (3y-1)^2+4=0\\ (3y-1)^2=-4\\ y\in \emptyset\)
\(9y^2 - 6y + 5 = 0 \\ \\a=9, \ b= -6 , \ c=5 \\ \\\Delta =b^2-4ac = (-6)^2 -4\cdot9\cdot 5 = 36 -180 = -144 \\ \\ If\ \Delta <0, \ then \ roots\ are \ imaginary \ (non-real)\)
