To solve the quadratic equation \( 10x^2 - 2x + 13 = 3 \), we first rearrange it into standard form:
\[ 10x^2 - 2x + 10 = 0 \]
Next, we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 10 \), \( b = -2 \), and \( c = 10 \). Plugging in these values:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 10 \cdot 10}}{2 \cdot 10} \]
Calculating under the square root:
\[ (-2)^2 = 4 \quad \text{and} \quad 4 \cdot 10 \cdot 10 = 400 \] \[ b^2 - 4ac = 4 - 400 = -396 \]
Now substitute back into the formula:
\[ x = \frac{2 \pm \sqrt{-396}}{20} \]
Since \(\sqrt{-396} = i\sqrt{396}\), we can simplify \(\sqrt{396}\):
\[ 396 = 4 \cdot 99 = 4 \cdot 9 \cdot 11 = 36 \cdot 11 = 6\sqrt{11} \] Thus, \(\sqrt{-396} = i \cdot 6\sqrt{11}\).
Now putting this all together:
\[ x = \frac{2 \pm 6i\sqrt{11}}{20} \] \[ = \frac{2}{20} \pm \frac{6i\sqrt{11}}{20} \] \[ = \frac{1}{10} \pm \frac{3i\sqrt{11}}{10} \]
So the solutions can be expressed in the form \( a \pm bi \):
\[ x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i \]
Now we can compare it with the options given:
- \( \frac{1}{10} \pm \frac{3\sqrt{11}}{10}i \)
This option matches:
Answer: \( \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \) (which corresponds to the option \( \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \) provided in your responses).