Question What are the roots of the equation? \[ 15 x^{2}-5 x+6=9 x^{2} \] Answer \( \frac{-514 \sqrt{119}}{12} \) subumit 3 and -2 \( \frac{54 \sqrt{119}}{12} \) \( \frac{3}{3} \) and \( -\frac{13}{2} \)

**Step 1. Write the equation in standard quadratic form.** We start with the given equation: \[ 15x^2 - 5x + 6 = 9x^2. \] Subtract \(9x^2\) from both sides: \[ 15x^2 - 9x^2 - 5x + 6 = 0. \] This simplifies to: \[ 6x^2 - 5x + 6 = 0. \] **Step 2. Compute the discriminant.** For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is: \[ \Delta = b^2 - 4ac. \] Here, \(a = 6\), \(b = -5\), and \(c = 6\). Thus, \[ \Delta = (-5)^2 - 4 \cdot 6 \cdot 6 = 25 - 144 = -119. \] **Step 3. Find the roots using the quadratic formula.** The quadratic formula is: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a}. \] Substitute the values we have: \[ x = \frac{-(-5) \pm \sqrt{-119}}{2 \cdot 6} = \frac{5 \pm \sqrt{-119}}{12}. \] Since \(\sqrt{-119} = i\sqrt{119}\), the roots become: \[ x = \frac{5 \pm i\sqrt{119}}{12}. \] **Final Answer:** The roots of the equation are \[ \boxed{\frac{5 \pm i\sqrt{119}}{12}}. \]