1. **State the problem:** We need to find the coordinates of the midpoint of points $A$ and $B$, where $A$ and $B$ are the intersection points of the curve $y = (10x - 3)(x+1)$ and the line $y - 6x = 0$. 2. **Rewrite the line equation:** The line equation $y - 6x = 0$ can be rewritten as $y = 6x$. 3. **Set the curve equal to the line to find intersection points:** Since both represent $y$, set $$ (10x - 3)(x+1) = 6x $$ 4. **Expand the left side:** $$ (10x - 3)(x+1) = 10x^2 + 10x - 3x - 3 = 10x^2 + 7x - 3 $$ 5. **Form the equation:** $$ 10x^2 + 7x - 3 = 6x $$ 6. **Bring all terms to one side:** $$ 10x^2 + 7x - 3 - 6x = 0 \implies 10x^2 + x - 3 = 0 $$ 7. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=10$, $b=1$, $c=-3$. Calculate the discriminant: $$ \Delta = 1^2 - 4 \times 10 \times (-3) = 1 + 120 = 121 $$ Calculate roots: $$ x = \frac{-1 \pm \sqrt{121}}{2 \times 10} = \frac{-1 \pm 11}{20} $$ So, $$ x_1 = \frac{-1 + 11}{20} = \frac{10}{20} = 0.5 $$ $$ x_2 = \frac{-1 - 11}{20} = \frac{-12}{20} = -0.6 $$ 8. **Find corresponding $y$ values using $y=6x$:** $$ y_1 = 6 \times 0.5 = 3 $$ $$ y_2 = 6 \times (-0.6) = -3.6 $$ 9. **Coordinates of points $A$ and $B$ are:** $$ A = (0.5, 3), \quad B = (-0.6, -3.6) $$ 10. **Find midpoint $M$ coordinates:** Midpoint formula: $$ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$ Calculate: $$ x_M = \frac{0.5 + (-0.6)}{2} = \frac{-0.1}{2} = -0.05 $$ $$ y_M = \frac{3 + (-3.6)}{2} = \frac{-0.6}{2} = -0.3 $$ **Final answer:** $$ \boxed{\left(-0.05, -0.3\right)} $$