To solve the quadratic equation \(10x^2 - 2x + 13 = 3\), we first move all terms to one side of the equation:
\[ 10x^2 - 2x + 13 - 3 = 0 \]
This simplifies to:
\[ 10x^2 - 2x + 10 = 0 \]
Next, we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 10\), \(b = -2\), and \(c = 10\). We can now calculate the discriminant (\(D\)):
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 10 \cdot 10 = 4 - 400 = -396 \]
Since the discriminant is negative, this indicates that the solutions will involve imaginary numbers. Now we calculate the solutions using the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20} \]
We can simplify \(\sqrt{-396}\):
\[ \sqrt{-396} = \sqrt{396} \cdot i = \sqrt{4 \cdot 99} \cdot i = 2\sqrt{99} \cdot i = 2 \cdot 3\sqrt{11} \cdot i = 6\sqrt{11} \cdot i \]
Now, substitute this back into the equation for \(x\):
\[ x = \frac{2 \pm 6\sqrt{11}i}{20} = \frac{1 \pm 3\sqrt{11}i}{10} \]
Thus, the solutions can be expressed as:
\[ x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i \]
The final answer in the form \(a \pm bi\) is:
\[ \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i \]
Therefore, the correct response from your options is:
\[ \text{Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i} \]
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