Algebra

The value for a can't be positive for the solutions for that equation. Since the roots are literally $\frac{-1\pm\sqrt{13}i}{2}$. b > 0 but not a.

Had there not been any limitations; we could've done the following:

First, we can start with some basic rearrangements.

$2x^2 - 10x + 13 = -6x^2 - 18x - 15$,

$8x^2 + 8x + 28 = 0$, 

$x^2 + x + \frac{7}{2} = 0$

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We turn this into a monic quadratic to simplify our working out. Since the roots of the quadratic are $a+bi, a-bi$

, we can use Vieta's formula to see that:

$(a+bi)+(a-bi) = 2a = -1 \rightarrow a = -\frac{1}{2}$
$(a+bi)\cdot(a-bi) = a^2 + b^2 = \frac{7}{2}$

We can substitute our value for a and get $a^2 + b^2 = \frac{7}{2} \rightarrow (-\frac{1}{2})^2 + b^2 = \frac{7}{2} \rightarrow b^2 = \frac{13}{4} \therefore b = \frac{\pm\sqrt{13}}{2}.$

Which means $a\cdot b = -\frac{1}{2}\cdot\frac{\pm\sqrt{13}}{2} = \frac{\pm\sqrt{13}}{4}.$