The zeros of [tex]x^2+3x+30[/tex] can be obtained by the quadratic formula (we use this method because the function is not factorable and has no real solutions, so it is easiest).
For a general quadratic [tex]ax^2+bx+c=0[/tex],
[tex]x= \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} [/tex].
For our function, a = 1, b = 3, and c = 30.
We substitute and solve for x:
[tex] =\dfrac{-(3) \pm \sqrt{(3)^2-4(1)(30)} }{2(1)}= \dfrac{-3 \pm \sqrt{9-120} }{2}\\\\\\= \dfrac{-3 \pm \sqrt{-111} }{2}=\dfrac{-3 \pm i\sqrt{111} }{2}[/tex]
[tex]x=\dfrac{-3+i \sqrt{111}}{2} \ \\\\\\\ x = \dfrac{-3-i \sqrt{111}}{2} \ [/tex]
The modulus of each complex solution is about 5.477, so neither solution is bigger or smaller.
