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Use the quadratic formula to solve. Express your answer in simplest form.
3, y, squared, minus, 10, y, minus, 10, equals, 3, y
3y
2
−10y−10=
3y

Answer 1

To solve the equation \( 3y^2 - 10y - 10 = 3y \) using the quadratic formula, we first need to rearrange the equation into standard form \( ax^2 + bx + c = 0 \).

1. Start with the original equation:
\[
3y^2 - 10y - 10 = 3y
\]

2. Move \( 3y \) to the left side of the equation:
\[
3y^2 - 10y - 3y - 10 = 0
\]

3. Combine like terms:
\[
3y^2 - 13y - 10 = 0
\]

Now, we can identify \( a \), \( b \), and \( c \) from the equation \( 3y^2 - 13y - 10 = 0 \):
- \( a = 3 \)
- \( b = -13 \)
- \( c = -10 \)

Next, we will use the quadratic formula, which is given by:
\[
y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]

4. Calculate \( b^2 - 4ac \):
\[
b^2 = (-13)^2 = 169
\]
\[
4ac = 4 \cdot 3 \cdot (-10) = -120
\]
\[
b^2 - 4ac = 169 - (-120) = 169 + 120 = 289
\]

Since \( 289 \) is a perfect square, we can proceed.

5. Calculate \( y \) using the quadratic formula:
\[
y = \frac{{-(-13) \pm \sqrt{289}}}{2 \cdot 3} = \frac{{13 \pm 17}}{6}
\]

Now we compute the two possible values for \( y \).

6. First value:
\[
y = \frac{{13 + 17}}{6} = \frac{30}{6} = 5
\]

7. Second value:
\[
y = \frac{{13 - 17}}{6} = \frac{-4}{6} = -\frac{2}{3}
\]

8. Thus, the solutions are:
\[
y = 5 \quad \text{and} \quad y = -\frac{2}{3}
\]

In simplest form, the solutions to the equation \( 3y^2 - 10y - 10 = 3y \) are:
\[
\boxed{5} \quad \text{and} \quad \boxed{-\frac{2}{3}}
\]