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

Step 1 ：Given the equation \(i^{2}+2 i+9=0\). This is a quadratic equation in the complex number \(i\). The general form of a quadratic equation is \(ax^2 + bx + c = 0\). 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 \(i^{2}+2 i+9=0\) are \(\boxed{-1 - 2.8284271247461903j}\) and \(\boxed{-1 + 2.8284271247461903j}\).