To solve the equation \(2x^2 = 6x - 5\) using the quadratic formula, we first need to rearrange the equation into standard form, which is \(ax^2 + bx + c = 0\).
Starting with the original equation:
\[ 2x^2 - 6x + 5 = 0 \]
Here, we have:
- \(a = 2\)
- \(b = -6\)
- \(c = 5\)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Now, we will substitute the values of \(a\), \(b\), and \(c\) into the formula.
First, calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = (-6)^2 = 36 \] \[ 4ac = 4 \cdot 2 \cdot 5 = 40 \] \[ b^2 - 4ac = 36 - 40 = -4 \]
Since the discriminant is negative, this means we will have complex solutions.
Now we will apply the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{-4}}{2 \cdot 2} \] \[ x = \frac{6 \pm \sqrt{-4}}{4} \]
Recall that \(\sqrt{-4}\) can be expressed as \(2i\):
\[ x = \frac{6 \pm 2i}{4} \]
Now simplify:
\[ x = \frac{6}{4} \pm \frac{2i}{4} \] \[ x = \frac{3}{2} \pm \frac{i}{2} \]
Thus, the solutions to the equation \(2x^2 = 6x - 5\) are:
\[ x = \frac{3}{2} + \frac{i}{2} \quad \text{and} \quad x = \frac{3}{2} - \frac{i}{2} \]
So, the final answers can be written as:
\[ x = \frac{3}{2} \pm \frac{i}{2} \]