Question

Use the quadratic formula to solve. Express your answer in simplest form.
15, r, squared, minus, 16, r, minus, 20, equals, minus, 5
15r
2
−16r−20=
−5

Answer 1

To solve the equation \( 15r^2 - 16r - 20 = -5 \) using the quadratic formula, we first need to rewrite the equation in standard form \( ax^2 + bx + c = 0 \).

Start by moving \(-5\) to the left side of the equation:

\[
15r^2 - 16r - 20 + 5 = 0
\]

This simplifies to:

\[
15r^2 - 16r - 15 = 0
\]

Now, identify the coefficients:

- \( a = 15 \)
- \( b = -16 \)
- \( c = -15 \)

Next, we apply the quadratic formula:

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

Substituting the values for \( a \), \( b \), and \( c \):

\[
r = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 15 \cdot (-15)}}{2 \cdot 15}
\]

Simplifying this step-by-step:

1. Calculate \( -b \):

\[
-r = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 15 \cdot (-15)}}{30}
\]

2. Calculate \( (-16)^2 \):

\[
(-16)^2 = 256
\]

3. Calculate \( 4ac \):

\[
4 \cdot 15 \cdot (-15) = -900
\]

So,

\[
b^2 - 4ac = 256 + 900 = 1156
\]

4. Take the square root of \( 1156 \):

\[
\sqrt{1156} = 34
\]

Now substitute this back into the quadratic formula:

\[
r = \frac{16 \pm 34}{30}
\]

Now we have two potential solutions:

1. For the plus case:

\[
r = \frac{16 + 34}{30} = \frac{50}{30} = \frac{5}{3}
\]

2. For the minus case:

\[
r = \frac{16 - 34}{30} = \frac{-18}{30} = \frac{-3}{5}
\]

Thus, the two solutions are:

\[
r = \frac{5}{3} \quad \text{and} \quad r = -\frac{3}{5}
\]

In simplest form, the final answers are:

\[
r = \frac{5}{3} \quad \text{and} \quad r = -\frac{3}{5}
\]