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 will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 10 \), \( b = -2 \), and \( c = 10 \).
Now, calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (-2)^2 = 4 \] \[ 4ac = 4 \cdot 10 \cdot 10 = 400 \] \[ b^2 - 4ac = 4 - 400 = -396 \]
Since the discriminant is negative, we will have complex solutions.
Now substituting these values into the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20} \]
Simplifying \( \sqrt{-396} \):
\[ \sqrt{-396} = \sqrt{396} i \]
Now we factor \( 396 \):
\[ 396 = 4 \cdot 99 = 4 \cdot 9 \cdot 11 = 36 \cdot 11 \]
Thus, \( \sqrt{396} = 6\sqrt{11} \):
\[ \sqrt{-396} = 6\sqrt{11} i \]
Therefore, substituting this back into the formula gives:
\[ x = \frac{2 \pm 6\sqrt{11}i}{20} \]
This simplifies as follows:
\[ x = \frac{2}{20} \pm \frac{6\sqrt{11}}{20}i = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]
Thus, the solutions to the quadratic equation \( 10x^2 - 2x + 13 = 3 \) in the form \( a \pm bi \) are:
\[ \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]
So, the correct answer is:
\[ \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]