Question

Solve the quadratic equation 3x2+3x+12=10 . Express its solutions in the form a±bi .

Answer 1

To solve the quadratic equation \( 3x^2 + 3x + 12 = 10 \), we first rearrange it into standard form:

\[
3x^2 + 3x + 12 - 10 = 0
\]
\[
3x^2 + 3x + 2 = 0
\]

Next, we can simplify this equation by dividing all terms by 3:

\[
x^2 + x + \frac{2}{3} = 0
\]

Now, to solve for \(x\), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = 1\), and \(c = \frac{2}{3}\).

First, we calculate the discriminant:

\[
b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot \frac{2}{3} = 1 - \frac{8}{3} = \frac{3}{3} - \frac{8}{3} = -\frac{5}{3}
\]

Since the discriminant is negative, we will get complex solutions. Now we plug in the values:

\[
x = \frac{-1 \pm \sqrt{-\frac{5}{3}}}{2 \cdot 1}
\]
\[
x = \frac{-1 \pm i\sqrt{\frac{5}{3}}}{2}
\]

This can be simplified further:

\[
x = \frac{-1}{2} \pm \frac{\sqrt{5}}{2\sqrt{3}} i
\]
\[
x = \frac{-1}{2} \pm \frac{\sqrt{15}}{6} i
\]

Thus, the solutions to the equation \( 3x^2 + 3x + 12 = 10 \) are:

\[
x = -\frac{1}{2} \pm \frac{\sqrt{15}}{6} i
\]

So the final answer can be express as:

\[
-\frac{1}{2} \pm \frac{\sqrt{15}}{6} i
\]