## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

To determine the number and type of solutions, we use the discriminant: $$x^2=\frac{1}{2}x-\frac{3}{5}\\ 10x^2-5x+6=0\\ \sqrt{\left(-5\right)^2-4\cdot \:10\cdot \:6} \\ \sqrt{-215}$$ Since the discriminant is negative, we see that there will be two imaginary solutions.