1. **State the problem:** Find the discriminant of the quadratic equation and determine how many real solutions it has. Given equation: $$6x = -x^2 + 1$$ 2. **Rewrite the equation in standard form:** Move all terms to one side to get $$ax^2 + bx + c = 0$$. $$-x^2 - 6x + 1 = 0$$ Multiply both sides by -1 to simplify: $$x^2 + 6x - 1 = 0$$ Here, $$a = 1$$, $$b = 6$$, and $$c = -1$$. 3. **Recall the discriminant formula:** $$\Delta = b^2 - 4ac$$ The discriminant tells us the nature of the roots: - If $$\Delta > 0$$, two distinct real solutions. - If $$\Delta = 0$$, one real solution (repeated root). - If $$\Delta < 0$$, no real solutions. 4. **Calculate the discriminant:** $$\Delta = 6^2 - 4(1)(-1) = 36 + 4 = 40$$ 5. **Interpret the result:** Since $$\Delta = 40 > 0$$, the equation has two distinct real solutions. **Final answer:** The discriminant is $$40$$, so the equation has two real solutions.