a) $i^2+2i+9=0$

Step 1 ：Given the equation $i^2+2i+9=0$. This is a quadratic equation in the complex number $i$. The general form of a quadratic equation is $a{x}^2+bx+c=0$. In this case, $a=1$, $b=2$, and $c=9$.

Step 2 ：We can solve this equation using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Step 3 ：Substitute $a=1$, $b=2$, and $c=9$ into the formula, we get $D=5.656854249492381j$.

Step 4 ：Then we find the solutions $sol1=(-1-2.8284271247461903j)$ and $sol2=(-1+2.8284271247461903j)$.

Step 5 ：These are complex numbers, where $j$ is the imaginary unit with the property that $j^2=-1$.

Step 6 ：Final Answer: The solutions to the equation $i^2+2i+9=0$ are $\boxed{{\displaystyle -1-2.8284271247461903j}}$ and $\boxed{{\displaystyle -1+2.8284271247461903j}}$.