Question

Find all solutions for x.
2x^2 + 2x + 5 = 0

Answer 1

To solve the equation \(2x^2 + 2x + 5 = 0\), we can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, the coefficients are:
- \(a = 2\)
- \(b = 2\)
- \(c = 5\)

First, we calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = 2^2 - 4 \cdot 2 \cdot 5 = 4 - 40 = -36
\]

Since the discriminant is negative (\(-36\)), this indicates that there are no real solutions, but there are two complex solutions.

Now we use the quadratic formula to find the complex solutions:

\[
x = \frac{-2 \pm \sqrt{-36}}{2 \cdot 2}
\]

Calculating the square root of \(-36\):

\[
\sqrt{-36} = 6i
\]

Now substitute this back into the equation:

\[
x = \frac{-2 \pm 6i}{4}
\]

This simplifies to:

\[
x = \frac{-2}{4} \pm \frac{6i}{4} = -\frac{1}{2} \pm \frac{3i}{2}
\]

Thus, the solutions are:

\[
x = -\frac{1}{2} + \frac{3i}{2} \quad \text{and} \quad x = -\frac{1}{2} - \frac{3i}{2}
\]

In summary, the solutions to the equation \(2x^2 + 2x + 5 = 0\) are:

\[
x = -\frac{1}{2} + \frac{3i}{2}, \quad x = -\frac{1}{2} - \frac{3i}{2}
\]