[tex]5x^2-3x+3=0\\ \\a=5 , \ \ b=-3, \ \ c=3\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{ 3-\sqrt{(-3)^2-4 \cdot 5\cdot3 }}{2 \cdot 5}=\frac{ 3-\sqrt{9-60}}{ 10}=\frac{ 3-\sqrt{-51}}{4}=\frac{ 3-\sqrt{ 51}i}{4} \\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= =\frac{ 3+\sqrt{(-3)^2-4 \cdot 5\cdot3 }}{2 \cdot 5}=\frac{ 3+\sqrt{9-60}}{ 10}=\frac{ 3+\sqrt{-51}}{10}=\frac{ 3+\sqrt{ 51}i}{10}\\ \\Answer: \ If b^2-4ac < 0, \ then \ roots \ are \ imaginary \ (non-real) [/tex]
