Using the quadratic formula, we get: \({-b \pm \sqrt{b^2-4ac} \over 2a}\Leftrightarrow\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^24\cdot \:1\cdot \:5}}{2\cdot \:1}\Leftrightarrow\frac{6\pm\sqrt{\left(-6\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1} \Leftrightarrow\frac{6\pm\sqrt{16}}{2\cdot \:1}\Leftrightarrow\frac{6\pm4}{2}=\boxed{5}, \boxed{1}.\)
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