To solve the quadratic equation \( 10x^2 - 2x + 13 = 3 \), we first rearrange it into standard form:
\[ 10x^2 - 2x + 13 - 3 = 0 \]
This simplifies to:
\[ 10x^2 - 2x + 10 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 10 \), \( b = -2 \), and \( c = 10 \). Plugging in these values:
- Calculate \( b^2 - 4ac \):
\[ b^2 = (-2)^2 = 4 \] \[ 4ac = 4 \cdot 10 \cdot 10 = 400 \] \[ b^2 - 4ac = 4 - 400 = -396 \]
- Now plug this into the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20} \]
- Simplify \( \sqrt{-396} \):
\[ \sqrt{-396} = \sqrt{396} i = \sqrt{4 \cdot 99} i = 2\sqrt{99} i = 2\sqrt{9 \cdot 11} i = 6\sqrt{11} i \]
- Substituting back gives:
\[ x = \frac{2 \pm 6\sqrt{11} i}{20} = \frac{2}{20} \pm \frac{6\sqrt{11} i}{20} = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]
Thus, the solutions of the equation are:
\[ x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]
Looking at the provided options, we can see that the correct expression for the solutions in the form \( a \pm bi \) is:
\[ \text{Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i} \]
Therefore, this is the answer.