## Algebra: A Combined Approach (4th Edition)

Since the discriminant ($b^2-4ac$) is equal to 0, the quadratic equations has one real solution: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} = \dfrac{-b\pm \sqrt{0}}{2a} = \dfrac{-b\pm 0}{2a} = -\dfrac{b}{2a}$