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.