To find the solutions to the quadratic equation \(2x^2 - 6x + 7 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \(2x^2 - 6x + 7 = 0\), the coefficients are:
- \(a = 2\)
- \(b = -6\)
- \(c = 7\)
Now we can substitute these values into the quadratic formula:
- Calculate the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4(2)(7) = 36 - 56 = -20 \]
Since the discriminant is negative, this indicates the solutions will be complex.
- Now substitute into the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{-20}}{2(2)} = \frac{6 \pm \sqrt{-20}}{4} \]
We can simplify \(\sqrt{-20}\):
\[ \sqrt{-20} = \sqrt{20} i = \sqrt{4 \cdot 5} i = 2\sqrt{5} i \]
- So now we substitute this back in:
\[ x = \frac{6 \pm 2\sqrt{5} i}{4} = \frac{6}{4} \pm \frac{2\sqrt{5} i}{4} = \frac{3}{2} \pm \frac{\sqrt{5} i}{2} \]
Thus, the solutions to the quadratic equation \(2x^2 - 6x + 7 = 0\) are:
\[ \frac{3}{2} \pm \frac{\sqrt{5}}{2} i \]
This corresponds to the answer:
\(32 \pm i \sqrt{5} / 2\) (Start Fraction 3 over 2 End Fraction plus or minus Start Fraction i Start Root 5 End Root over 2 End Fraction)
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