To solve the quadratic equation \(2x^2 = 4x - 7\), we need to first rewrite it in the standard form \(ax^2 + bx + c = 0\).
1. Rewrite the equation:
[tex]\[ 2x^2 - 4x + 7 = 0 \][/tex]
Here, \(a = 2\), \(b = -4\), and \(c = 7\).
2. Calculate the discriminant \(\Delta\) using the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of \(a\), \(b\), and \(c\):
[tex]\[ \Delta = (-4)^2 - 4 \cdot 2 \cdot 7 \][/tex]
[tex]\[ \Delta = 16 - 56 \][/tex]
[tex]\[ \Delta = -40 \][/tex]
3. Since the discriminant is negative (\(\Delta = -40\)), the solutions to the quadratic equation will be complex numbers. We use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
4. Substitute \(b = -4\), \(a = 2\), and \(\Delta = -40\) into the quadratic formula:
[tex]\[ x = \frac{-(-4) \pm \sqrt{-40}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{-40}}{4} \][/tex]
5. Simplify \(\sqrt{-40}\):
[tex]\[ \sqrt{-40} = \sqrt{-1 \cdot 40} = \sqrt{-1} \cdot \sqrt{40} = i \cdot \sqrt{40} \][/tex]
Since \(\sqrt{40}\) can be further simplified:
[tex]\[ \sqrt{40} = \sqrt{4 \cdot 10} = \sqrt{4} \cdot \sqrt{10} = 2\sqrt{10} \][/tex]
So,
[tex]\[ \sqrt{-40} = 2i\sqrt{10} \][/tex]
6. Substitute back to the formula:
[tex]\[ x = \frac{4 \pm 2i\sqrt{10}}{4} \][/tex]
[tex]\[ x = \frac{4}{4} \pm \frac{2i\sqrt{10}}{4} \][/tex]
[tex]\[ x = 1 \pm \frac{i\sqrt{10}}{2} \][/tex]
Therefore, the solutions to the quadratic equation \(2x^2 = 4x - 7\) are:
[tex]\[ x = 1 \pm \frac{i\sqrt{10}}{2} \][/tex]
Comparing with the given options, this matches:
[tex]\[ \frac{2 \pm \sqrt{10}i}{2} \][/tex]
Thus, the answer is:
[tex]\[ \frac{2 \pm \sqrt{10} i}{2} \][/tex]
