Question

Use the quadratic formula to solve. Express your answer in simplest form.
12, b, squared, plus, 21, b, plus, 10, equals, 3, b, squared
12b
2
+21b+10=
3b
2

Answer 1

To solve the quadratic equation \(12b^2 + 21b + 10 = 3b^2\), we first rearrange it into standard form \(ax^2 + bx + c = 0\).

1. Move \(3b^2\) to the left side of the equation:
\[
12b^2 - 3b^2 + 21b + 10 = 0
\]
Simplifying this, we get:
\[
9b^2 + 21b + 10 = 0
\]

2. The quadratic equation is now in the form \(9b^2 + 21b + 10 = 0\), where \(a = 9\), \(b = 21\), and \(c = 10\).

3. We can apply the quadratic formula, which is given by:
\[
b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
Here \(A = 9\), \(B = 21\), and \(C = 10\).

4. First, calculate the discriminant \(B^2 - 4AC\):
\[
B^2 = 21^2 = 441
\]
\[
4AC = 4 \cdot 9 \cdot 10 = 360
\]
So:
\[
B^2 - 4AC = 441 - 360 = 81
\]

5. Since the discriminant is positive, we have two real solutions. Now, substituting in the quadratic formula:
\[
b = \frac{-21 \pm \sqrt{81}}{2 \cdot 9}
\]

6. Calculate \(\sqrt{81}\):
\[
\sqrt{81} = 9
\]

7. Substitute \(\sqrt{81}\) back into the formula:
\[
b = \frac{-21 \pm 9}{18}
\]

8. Now we calculate the two possible solutions:
- For the positive case:
\[
b = \frac{-21 + 9}{18} = \frac{-12}{18} = -\frac{2}{3}
\]
- For the negative case:
\[
b = \frac{-21 - 9}{18} = \frac{-30}{18} = -\frac{5}{3}
\]

Thus, the solutions to the quadratic equation \(9b^2 + 21b + 10 = 0\) are:
\[
b = -\frac{2}{3} \quad \text{and} \quad b = -\frac{5}{3}
\]