For what values of a does the equation (a^2 + 2a)x^2 + (3a + 5)x+2 = 0 yield no real solutions x? Express your answer in interval notation.
(a2 + 2a)x2 + (3a + 5)x + 2 = 0
For imaginary solutions, b2 - 4ac < 0
(3a + 5)2 - 4(a2 + 2a)2 < 0
9a2 + 30a + 25 - 8a2 - 16a < 0
a2 + 14a + 25 < 0 => \(a = {-7 \pm 2sqrt{6} }\)
\(a = {-14 \pm \sqrt{96} \over 2}\)
