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The image you uploaded contains five quadratic equations that need to be solved. I will solve each equation step-by-step.

### 1. \( 4x^2 + 8x + 3 = 0 \)

To solve this quadratic equation, we'll use the quadratic formula:

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

Here, \(a = 4\), \(b = 8\), and \(c = 3\).

\[
x = \frac{-8 \pm \sqrt{8^2 - 4(4)(3)}}{2(4)}
\]
\[
x = \frac{-8 \pm \sqrt{64 - 48}}{8}
\]
\[
x = \frac{-8 \pm \sqrt{16}}{8}
\]
\[
x = \frac{-8 \pm 4}{8}
\]

So, the solutions are:

\[
x_1 = \frac{-8 + 4}{8} = \frac{-4}{8} = -\frac{1}{2}
\]

\[
x_2 = \frac{-8 - 4}{8} = \frac{-12}{8} = -\frac{3}{2}
\]

### 2. \( 2x^2 + 3x - 20 = 0 \)

Using the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-20)}}{2(2)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 160}}{4}
\]
\[
x = \frac{-3 \pm \sqrt{169}}{4}
\]
\[
x = \frac{-3 \pm 13}{4}
\]

So, the solutions are:

\[
x_1 = \frac{-3 + 13}{4} = \frac{10}{4} = \frac{5}{2}
\]

\[
x_2 = \frac{-3 - 13}{4} = \frac{-16}{4} = -4
\]

### 3. \( 5x^2 - 80 = 0 \)

This equation can be simplified by dividing through by 5:

\[
x^2 - 16 = 0
\]

Factor it:

\[
(x - 4)(x + 4) = 0
\]

The solutions are:

\[
x_1 = 4
\]
\[
x_2 = -4
\]

### 4. \( 5x^2 + 9x = -4 \)

First, move everything to one side to set the equation to zero:

\[
5x^2 + 9x + 4 = 0
\]

Now, use the quadratic formula:

\[
x = \frac{-9 \pm \sqrt{9^2 - 4(5)(4)}}{2(5)}
\]
\[
x = \frac{-9 \pm \sqrt{81 - 80}}{10}
\]
\[
x = \frac{-9 \pm \sqrt{1}}{10}
\]
\[
x = \frac{-9 \pm 1}{10}
\]

So, the solutions are:

\[
x_1 = \frac{-9 + 1}{10} = \frac{-8}{10} = -\frac{4}{5}
\]

\[
x_2 = \frac{-9 - 1}{10} = \frac{-10}{10} = -1
\]

### 5. \( 9x^2 + 4x - 16 = 0 \)

Using the quadratic formula:

\[
x = \frac{-4 \pm \sqrt{4^2 - 4(9)(-16)}}{2(9)}
\]
\[
x = \frac{-4 \pm \sqrt{16 + 576}}{18}
\]
\[
x = \frac{-4 \pm \sqrt{592}}{18}
\]
\[
x = \frac{-4 \pm \sqrt{16 \times 37}}{18} = \frac{-4 \pm 4\sqrt{37}}{18}
\]
\[
x = \frac{-2 \pm 2\sqrt{37}}{9}
\]

So, the solutions are:

\[
x_1 = \frac{-2 + 2\sqrt{37}}{9}
\]
\[
x_2 = \frac{-2 - 2\sqrt{37}}{9}
\]

---

Would you like more details on any of these solutions? Here are five additional questions you might find interesting:

1. How can the discriminant help determine the nature of the roots for a quadratic equation?
2. How would you solve a quadratic equation by completing the square?
3. What is the significance of the coefficients in a quadratic equation?
4. How do you factorize a quadratic equation?
5. Can you apply the quadratic formula to non-real solutions?

**Tip:** The discriminant \(b^2 - 4ac\) is key in determining whether the roots of a quadratic equation are real and distinct, real and equal, or complex.

Solve the quadratic equations: 1. 4x^2 + 8x + 3 = 0, 2. 2x^2 + 3x - 20 = 0, 3. 5x^2 - 80 = 0, 4. 5x^2 + 9x = -4, 5. 9x^2 + 4x - 16 = 0

This page provides a detailed step-by-step solution to five quadratic equations, including 4x^2 + 8x + 3 = 0. Learn key algebraic concepts such as the quadratic formula, factoring, and simplification. Suitable for students in Grades 8-10.

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