Question

Solve the quadratic equation.
13+2x^2=10x

Answer 1

To solve the quadratic equation \( 13 + 2x^2 = 10x \), we first rearrange it into standard form:

\[
2x^2 - 10x + 13 = 0
\]

Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -10 \), and \( c = 13 \).

Now, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = (-10)^2 = 100
\]
\[
4ac = 4 \cdot 2 \cdot 13 = 104
\]
\[
b^2 - 4ac = 100 - 104 = -4
\]

Since the discriminant is negative (\(-4\)), this means there are no real solutions, but there are two complex solutions. Now we continue with the quadratic formula:

\[
x = \frac{-(-10) \pm \sqrt{-4}}{2 \cdot 2}
\]
\[
x = \frac{10 \pm \sqrt{-4}}{4}
\]

Since \(\sqrt{-4} = 2i\), we simply plug it in:

\[
x = \frac{10 \pm 2i}{4}
\]
\[
x = \frac{10}{4} \pm \frac{2i}{4}
\]
\[
x = \frac{5}{2} \pm \frac{i}{2}
\]

Therefore, the solutions to the equation \( 13 + 2x^2 = 10x \) are:

\[
x = \frac{5}{2} + \frac{i}{2} \quad \text{and} \quad x = \frac{5}{2} - \frac{i}{2}
\]