The quadratic 2x^2 - 3x + 21 has two imaginary roots. What is the sum of the squares of these roots? Express your answer as a decimal rounded to the nearest hundredth.
Let one root be a and the other be b. We seek to find a^2+b^2, which is (a+b)^2-2ab. By vietas, we have a+b=3/2, so (a+b)^2=9/4. We can find 2ab as 21/2*2=21. Then, we have 9/4-21=-75/4=-18.75
