To solve the equation \(2x^2 = 6x - 5\), we first rearrange it into standard quadratic form:
\[ 2x^2 - 6x + 5 = 0 \]
Next, we can use the quadratic formula to find the solutions. The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \(2x^2 - 6x + 5 = 0\), the coefficients are:
- \(a = 2\)
- \(b = -6\)
- \(c = 5\)
Let's calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4 \]
Since the discriminant is negative, we have two complex conjugate solutions. Now we can substitute into the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{-4}}{2(2)} = \frac{6 \pm \sqrt{-4}}{4} \]
We know that \(\sqrt{-4} = 2i\). Therefore, we have:
\[ x = \frac{6 \pm 2i}{4} \]
This simplifies to:
\[ x = \frac{6}{4} \pm \frac{2i}{4} = \frac{3}{2} \pm \frac{i}{2} \]
Thus, the solutions are:
\[ x = \frac{3}{2} \pm \frac{i}{2} \]
So, the final answer is:
\[ x = \frac{3}{2} \pm \frac{i}{2} \]
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