\(6b ^ 2-17b +10 \\ \\\Delta = b^{2}-4ac = (-17)^{2}-4*6*10=289-240=49\\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{17- \sqrt{49}}{2*6}=\frac{17-7}{12}= \frac{10}{12}=\frac{5}{6}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{17+ \sqrt{49}}{2*6}=\frac{17+7}{12}= \frac{24}{12}=2 \\ \\ Answer : \ x=\frac{5}{6} \ \ or \ \ x=2\)
