Question

solve for x using quadratic formula
2x^2-3x5=0

Answer 1

It looks like there might be a small typo in your equation. I'll assume you meant to write \(2x^2 - 3x + 5 = 0\).

To solve the quadratic equation \(ax^2 + bx + c = 0\) using the quadratic formula:

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

Here, \(a = 2\), \(b = -3\), and \(c = 5\).

Let's plug these values into the formula:

1. Calculate \(b^2 - 4ac\):

\[
b^2 = (-3)^2 = 9
\]
\[
4ac = 4 \times 2 \times 5 = 40
\]
\[
b^2 - 4ac = 9 - 40 = -31
\]

Since the discriminant (the part under the square root) is negative (\(-31\)), this means the equation has no real solutions, but two complex solutions.

2. Now, substitute into the quadratic formula:

\[
x = \frac{{-(-3) \pm \sqrt{-31}}}{{2 \times 2}}
\]
\[
x = \frac{{3 \pm \sqrt{-31}}}{4}
\]

3. Express the complex solutions:

The square root of \(-31\) can be written as \(i\sqrt{31}\), where \(i\) is the imaginary unit. So the solutions can be expressed as:

\[
x = \frac{3 \pm i\sqrt{31}}{4}
\]

Thus, the final solutions for \(x\) are:

\[
x = \frac{3}{4} + \frac{i\sqrt{31}}{4}, \quad x = \frac{3}{4} - \frac{i\sqrt{31}}{4}
\]