Question

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

Answer 1

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

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

This simplifies to:

\[
3x^2 + 3x + 2 = 0
\]

Next, we will use the quadratic formula, which is given by:

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

For our equation, we identify \( a = 3 \), \( b = 3 \), and \( c = 2 \). We will first calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 3^2 - 4(3)(2) = 9 - 24 = -15
\]

Since the discriminant is negative, the solutions will be complex. We can now apply the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{-15}}{2(3)}
\]

This simplifies to:

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

We can further simplify this:

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

Which gives:

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

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

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

This is the final answer in the form \( a \pm bi \).