1. The problem states that we have constants $c=3$, $a=1$, and $b=11.5$. 2. Typically, these variables might be part of an equation such as a quadratic $ax^2 + bx + c = 0$. 3. The quadratic formula to find roots is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 4. Substitute the values: $$x = \frac{-11.5 \pm \sqrt{(11.5)^2 - 4 \times 1 \times 3}}{2 \times 1}$$. 5. Calculate the discriminant: $$11.5^2 = 132.25$$ and $$4 \times 1 \times 3 = 12$$, so discriminant $$= 132.25 - 12 = 120.25$$. 6. Find the square root: $$\sqrt{120.25} = 10.97$$ (approx). 7. Calculate the two roots: - $$x_1 = \frac{-11.5 + 10.97}{2} = \frac{-0.53}{2} = -0.265$$ (approx) - $$x_2 = \frac{-11.5 - 10.97}{2} = \frac{-22.47}{2} = -11.235$$ (approx) 8. Therefore, the solutions to the quadratic equation are approximately $$x = -0.265$$ and $$x = -11.235$$.