https://file.aiquickdraw.com/mathbot/file/1727058173969nz16f2l0.jpg

The equation in the image is:

\[
15x + 3 = 20x^2 + 4x
\]

To solve for \( x \), let's first rearrange the equation by moving all terms to one side:

\[
0 = 20x^2 + 4x - 15x - 3
\]

Simplifying:

\[
0 = 20x^2 - 11x - 3
\]

Now we can solve this quadratic equation using the quadratic formula. The general form of the quadratic formula is:

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

For the equation \( 20x^2 - 11x - 3 = 0 \), we have:

- \( a = 20 \)
- \( b = -11 \)
- \( c = -3 \)

Substituting these values into the quadratic formula:

\[
x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(20)(-3)}}{2(20)}
\]
\[
x = \frac{11 \pm \sqrt{121 + 240}}{40}
\]
\[
x = \frac{11 \pm \sqrt{361}}{40}
\]
\[
x = \frac{11 \pm 19}{40}
\]

Now, solving for the two possible values of \( x \):

1. \( x_1 = \frac{11 + 19}{40} = \frac{30}{40} = \frac{3}{4} \)
2. \( x_2 = \frac{11 - 19}{40} = \frac{-8}{40} = \frac{-1}{5} \)

Thus, the solutions are \( x = \frac{3}{4} \) and \( x = \frac{-1}{5} \).

### Answer: The solution(s) is/are \( x = \frac{3}{4}, -\frac{1}{5} \).

Would you like more details on how I derived this? Here are some related questions for further understanding:

1. How does the quadratic formula work in solving equations?
2. Can this equation be solved by factoring instead of using the quadratic formula?
3. How do you determine if a quadratic equation has no real solutions?
4. What are other methods to solve quadratic equations besides the quadratic formula?
5. How do you graph the solutions of a quadratic equation?

**Tip:** When solving quadratic equations, always check the discriminant \( b^2 - 4ac \) first. It tells you how many real solutions there are!

Use a graphing utility to find or approximate solution(s) of the given equation: 15x + 3 = 20x^2 + 4x.

Learn how to solve the quadratic equation 15x + 3 = 20x^2 + 4x by approximating solutions using graphing utility. This example demonstrates how to use the quadratic formula, solving for x with detailed steps. Ideal for students in Grades 9-12.

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