Solve the quadratic equation 10x2−2x+13=3

To solve the equation \( 10x^2 - 2x + 13 = 3 \), we first rearrange it to standard form.

1. Subtract 3 from both sides: \[ 10x^2 - 2x + 10 = 0 \]

2. Now, identify the coefficients for the quadratic equation \( ax^2 + bx + c = 0 \):

   • \( a = 10 \)
   • \( b = -2 \)
   • \( c = 10 \)

3. Use the quadratic formula to find the roots: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

4. Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 - 4ac = (-2)^2 - 4 \cdot 10 \cdot 10 = 4 - 400 = -396 \]

5. Since the discriminant is negative, the solutions will be complex numbers: \[ x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20} \]

6. Rewrite the square root of the negative number: \[ \sqrt{-396} = \sqrt{396} i \]

   • Factor \( 396 \) to simplify: \[ 396 = 4 \cdot 99 = 4 \cdot 9 \cdot 11 = 36 \cdot 11 = 6^2 \cdot 11 \]
   • Thus: \[ \sqrt{396} = 6\sqrt{11} \]

7. Substitute back into the equation: \[ x = \frac{2 \pm 6\sqrt{11} i}{20} = \frac{1 \pm 3\sqrt{11} i}{10} \]

Thus, the solutions are: \[ x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]

From the list of choices provided, the correct answer is: \[ \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i \]