Find all real solutions to the equation: \( 2 x^{3}-25=-x^{2}+50 x \) \( x=\square \)

To solve the equation: \[ 2x^{3} - 25 = -x^{2} + 50x \] First, move all terms to one side of the equation: \[ 2x^{3} + x^{2} - 50x - 25 = 0 \] We now have a cubic equation: \[ 2x^3 + x^2 - 50x - 25 = 0 \] **Step 1: Find Rational Roots** Using the Rational Root Theorem, possible rational roots are factors of the constant term divided by factors of the leading coefficient. The possible rational roots are: \[ \pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 25, \pm \frac{25}{2} \] Testing \( x = 5 \): \[ 2(5)^3 + (5)^2 - 50(5) - 25 = 250 + 25 - 250 - 25 = 0 \] So, \( x = 5 \) is a root. **Step 2: Factor the Polynomial** Factor out \( (x - 5) \) from the cubic equation: \[ 2x^3 + x^2 - 50x - 25 = (x - 5)(2x^2 + 11x + 5) \] **Step 3: Solve the Quadratic Equation** Now, solve the quadratic equation: \[ 2x^2 + 11x + 5 = 0 \] Using the quadratic formula: \[ x = \frac{-11 \pm \sqrt{121 - 40}}{4} = \frac{-11 \pm \sqrt{81}}{4} = \frac{-11 \pm 9}{4} \] This gives two solutions: \[ x = \frac{-11 + 9}{4} = -\frac{1}{2} \quad \text{and} \quad x = \frac{-11 - 9}{4} = -5 \] **Final Solutions** All real solutions to the equation are: \[ x = 5, \quad x = -5, \quad \text{and} \quad x = -\frac{1}{2} \] **Answer:** \(x = 5,\; x = -5,\; \text{and}\; x = -\dfrac{1}{2}\)