\begin{tabular}{l|l|lll} & Question \\ What are the roots of the equation? \\ & \( -3 x=-10 x^{2}-4 \)\end{tabular}

We start with the equation \[ -3x = -10x^2 -4. \] First, we bring all terms to one side to write the equation in standard quadratic form: \[ -3x + 10x^2 + 4 = 0. \] Reordering the terms we obtain: \[ 10x^2 - 3x + 4 = 0. \] 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, we have \( a = 10 \), \( b = -3 \), and \( c = 4 \). Plugging these into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot 10 \cdot 4}}{2\cdot 10}. \] Simplify step by step: \[ x = \frac{3 \pm \sqrt{9 - 160}}{20}, \] \[ x = \frac{3 \pm \sqrt{-151}}{20}. \] Since the discriminant \( 9 - 160 = -151 \) is negative, the equation has two complex roots. We can express the square root of the negative number using the imaginary unit \( i \): \[ \sqrt{-151} = i\sqrt{151}. \] Thus, the roots of the equation are: \[ x = \frac{3 \pm i\sqrt{151}}{20}. \]