1. The problem is to solve the quadratic equation $5x^2 - 11x + 6 = 0$. 2. We use the quadratic formula to find the roots of $ax^2 + bx + c = 0$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=-11$, and $c=6$. 3. Calculate the discriminant: $$b^2 - 4ac = (-11)^2 - 4 \times 5 \times 6 = 121 - 120 = 1$$ 4. Substitute values into the quadratic formula: $$x = \frac{-(-11) \pm \sqrt{1}}{2 \times 5} = \frac{11 \pm 1}{10}$$ 5. Find the two solutions: - For the plus sign: $$x = \frac{11 + 1}{10} = \frac{12}{10} = \frac{6}{5}$$ - For the minus sign: $$x = \frac{11 - 1}{10} = \frac{10}{10} = 1$$ 6. Therefore, the solutions are $x=1$ and $x=\frac{6}{5}$. 7. The answers $x=5$ and $x=6$ are incorrect because they do not satisfy the equation when substituted back. This explains why the real answers are $x=1$ and $x=\frac{6}{5}$, not $5$ and $6$.