Question

find all complex solutions of $2x^{2}+6x + 5 = 0$. (if there is more than one solution, separate them with commas.)

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 2\), \(b = 6\), \(c = 5\).

Step2: Calculate Discriminant

First, find the discriminant \(D=b^{2}-4ac\). Substitute the values: \(D = 6^{2}-4\times2\times5=36 - 40=- 4\).

Step3: Substitute into Quadratic Formula

Now, substitute \(a\), \(b\), and \(D\) into the quadratic formula. \(x=\frac{-6\pm\sqrt{-4}}{2\times2}\). Since \(\sqrt{-4}=\sqrt{4}\times\sqrt{-1} = 2i\) (where \(i\) is the imaginary unit with \(i^{2}=-1\)), we have \(x=\frac{-6\pm2i}{4}\).

Step4: Simplify the Expression

Simplify the fraction: \(x=\frac{-3\pm i}{2}\), which can be written as \(x=-\frac{3}{2}+\frac{1}{2}i,-\frac{3}{2}-\frac{1}{2}i\).

Answer:

\(-\frac{3}{2}+\frac{1}{2}i,-\frac{3}{2}-\frac{1}{2}i\)