What are the roots of the equation? \[ 3 x^{2}-16 x+15=3 \] Answer \( \begin{array}{l}\frac{2}{3} \text { and }-\frac{2}{3} \\ \frac{16 \pm \sqrt{112}}{6} \\ \frac{-16 \pm \sqrt{112}}{6} \\ \text { Submit } \\ \text { Answer }\end{array} \)

We start with the equation \[ 3x^2 - 16x + 15 = 3. \] **Step 1. Subtract 3 from both sides** Subtracting \(3\) from both sides gives: \[ 3x^2 - 16x + 15 - 3 = 0, \] which simplifies to \[ 3x^2 - 16x + 12 = 0. \] **Step 2. Apply the quadratic formula** For a quadratic equation of the form \[ ax^2 + bx + c = 0, \] the roots are given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Here, \(a = 3\), \(b = -16\), and \(c = 12\). Plugging these values into the formula gives: \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 12}}{2 \cdot 3}. \] **Step 3. Simplify the expression** Calculate the numerator: \[ -(-16) = 16. \] Compute the discriminant: \[ (-16)^2 - 4 \cdot 3 \cdot 12 = 256 - 144 = 112. \] Substitute these back into the equation: \[ x = \frac{16 \pm \sqrt{112}}{6}. \] **Step 4. Final answer** Thus, the roots of the equation are: \[ x = \frac{16 \pm \sqrt{112}}{6}. \]