To determine when the soccer ball will hit the ground, we need to set the height \( y \) equal to 0 and solve for \( t \) in the quadratic equation:

Question

\[
0 = -16t^2 + 20t + 3.
\]

This can be rearranged to the standard form of a quadratic equation:

\[
-16t^2 + 20t + 3 = 0.
\]

To make it easier to work with, we can multiply the entire equation by -1 to eliminate the negative leading coefficient:

\[
16t^2 - 20t - 3 = 0.
\]

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

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

where \( a = 16 \), \( b = -20 \), and \( c = -3 \).

### Step 1: Calculate the discriminant (\( b^2 - 4ac \))

First, compute \( b^2 \):

\[
b^2 = (-20)^2 = 400.
\]

Next, compute \( 4ac \):

\[
4ac = 4 \cdot 16 \cdot (-3) = -192.
\]

Now, calculate the discriminant:

\[
b^2 - 4ac = 400 - (-192) = 400 + 192 = 592.
\]

### Step 2: Apply the quadratic formula

Now substitute the values into the quadratic formula:

\[
t = \frac{-(-20) \pm \sqrt{592}}{2 \cdot 16},
\]

which simplifies to:

\[
t = \frac{20 \pm \sqrt{592}}{32}.
\]

### Step 3: Simplify \(\sqrt{592}\)

To simplify \(\sqrt{592}\):

\[
592 = 16 \cdot 37,
\]
so:

\[
\sqrt{592} = \sqrt{16 \cdot 37} = 4\sqrt{37}.
\]

### Step 4: Substitute back into the formula

Now we can write:

\[
t = \frac{20 \pm 4\sqrt{37}}{32}.
\]

This can be further simplified:

\[
t = \frac{20}{32} \pm \frac{4\sqrt{37}}{32} = \frac{5}{8} \pm \frac{\sqrt{37}}{8}.
\]

### Step 5: Calculate the two potential solutions

Thus, we have:

\[
t_1 = \frac{5 + \sqrt{37}}{8},
\]
\[
t_2 = \frac{5 - \sqrt{37}}{8}.
\]

### Step 6: Determine the positive solution

Since we're looking for a positive time, we consider:

#### Estimate \(\sqrt{37} \approx 6.08\)

Calculating:

1. For \( t_1 \):

\[
t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.385.
\]

2. For \( t_2 \):

\[
t_2 \approx \frac{5 - 6.08}{8} = \frac{-1.08}{8} \approx -0.135.
\]

Since time cannot be negative, we have:

\[
t \approx 1.385 \text{ seconds}.
\]

### Conclusion

The soccer ball will hit the ground approximately **1.39 seconds** after being kicked.

make this like a 7th grader wrote it

Answer 1

To find out when the soccer ball will hit the ground, we need to figure out when its height \( y \) is 0. We start with the equation:

\[ 0 = -16t^2 + 20t + 3. \]

This is a quadratic equation! To make it easier to work with, let's rearrange it to:

\[ -16t^2 + 20t + 3 = 0. \]

We can multiply everything by -1 to get rid of the negative sign:

\[ 16t^2 - 20t - 3 = 0. \]

Now, we can use the quadratic formula to find \( t \). The quadratic formula is:

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

where \( a = 16 \), \( b = -20 \), and \( c = -3 \).

Step 1: Calculate the discriminant

First, let's find \( b^2 \):

\[ b^2 = (-20)^2 = 400. \]

Next, we need to calculate \( 4ac \):

\[ 4ac = 4 \cdot 16 \cdot (-3) = -192. \]

Now we can find out what's called the discriminant:

\[ b^2 - 4ac = 400 - (-192) = 400 + 192 = 592. \]

Step 2: Use the quadratic formula

Now we put our numbers into the quadratic formula:

\[ t = \frac{-(-20) \pm \sqrt{592}}{2 \cdot 16}. \]

This simplifies to:

\[ t = \frac{20 \pm \sqrt{592}}{32}. \]

Step 3: Simplify \(\sqrt{592}\)

Next, we’ll simplify \(\sqrt{592}\):

\[ 592 = 16 \cdot 37, \] so:

\[ \sqrt{592} = \sqrt{16 \cdot 37} = 4\sqrt{37}. \]

Step 4: Substitute back into the formula

Now we have:

\[ t = \frac{20 \pm 4\sqrt{37}}{32}. \]

We can simplify this a bit more:

\[ t = \frac{20}{32} \pm \frac{4\sqrt{37}}{32} = \frac{5}{8} \pm \frac{\sqrt{37}}{8}. \]

Step 5: Calculate the two potential solutions

So, we have two different times:

\[ t_1 = \frac{5 + \sqrt{37}}{8}, \] \[ t_2 = \frac{5 - \sqrt{37}}{8}. \]

Step 6: Find the positive solution

Since time can’t be negative, let’s check which one is positive:

Estimating \(\sqrt{37} \approx 6.08\)

1. For \( t_1 \):

\[ t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.39. \]

2. For \( t_2 \):

\[ t_2 \approx \frac{5 - 6.08}{8} = \frac{-1.08}{8} \approx -0.135. \]

Since time can’t be negative, we ignore \( t_2 \).

Conclusion

The soccer ball will hit the ground in about 1.39 seconds after being kicked!