\begin{tabular}{l} hilliard.instructure.com \( \mathrm{Q}- \) Private \\ \hline Question 9 \\ Solve for all solutions of x : \\ \( x^{2}+2 x=-2 \) \\ Ox=\{1\} \\ \hline\end{tabular}

To solve the equation \( x^{2} + 2x + 2 = 0 \), we first rearrange it to standard quadratic form. Then, we can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 1, b = 2, c = 2 \). Plugging in these values gives us \( x = \frac{-2 \pm \sqrt{(2)^{2} - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} \). This results in complex solutions \( x = -1 \pm i \). Another interesting aspect of quadratic equations is that they can always provide a minimum or maximum value depending on the direction of the parabola. When you analyze the vertex of the quadratic, you can easily determine the range of the solutions. For this equation, as it's opening upwards (since \( a > 0 \)), the minimum point (the vertex) is at \( x = -1 \), meaning the function will have a minimum value, but all the solutions lie above that due to the complex roots.