Quadratic Equations: (y+1/4)2 = 9/16 (x+3)2 = 4.9 3x2−5x−2=0 (2x−1)2 = 1 x2+5x−2.4=0

$$y + \frac{1}{4} = \pm\sqrt{\frac{9}{16}}$$ $$y + \frac{1}{4} = \pm\frac{3}{4}$$

$$y = -\frac{1}{4} \pm \frac{3}{4}$$

This gives two solutions:

$$y_1 = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2}$$ $$y_2 = -\frac{1}{4} - \frac{3}{4} = -\frac{4}{4} = -1$$

The solutions are $\boxed{y = \frac{1}{2}, -1}$.

2) $(x + 3)^2 = 49$

$$x + 3 = \pm\sqrt{49}$$ $$x + 3 = \pm 7$$

$$x = -3 \pm 7$$

This gives two solutions:

$$x_1 = -3 + 7 = 4$$ $$x_2 = -3 - 7 = -10$$

The solutions are $\boxed{x = 4, -10}$.

3) $3x^2 - 5x - 2 = 0$

We look for two numbers that multiply to $3 \times -2 = -6$ and add to $-5$. These numbers are $-6$ and 1.

$$3x^2 - 6x + x - 2 = 0$$ $$3x(x - 2) + 1(x - 2) = 0$$ $$(3x + 1)(x - 2) = 0$$

$$3x + 1 = 0 \implies 3x = -1 \implies x_1 = -\frac{1}{3}$$ $$x - 2 = 0 \implies x_2 = 2$$

The solutions are $\boxed{x = -\frac{1}{3}, 2}$.

4) $(2x - 1)^2 = 1$

$$2x - 1 = \pm\sqrt{1}$$ $$2x - 1 = \pm 1$$

$$2x = 1 \pm 1$$

This gives two solutions:

$$2x_1 = 1 + 1 = 2 \implies x_1 = 1$$ $$2x_2 = 1 - 1 = 0 \implies x_2 = 0$$

The solutions are $\boxed{x = 1, 0}$.

5) $x^2 + 5x - 24 = 0$

We look for two numbers that multiply to $-24$ and add to 5. These numbers are 8 and $-3$.

$$(x + 8)(x - 3) = 0$$

$$x + 8 = 0 \implies x_1 = -8$$ $$x - 3 = 0 \implies x_2 = 3$$

The solutions are $\boxed{x = -8, 3}$.

For $x^2 = 16$ (from the top right of the board):

$$x = \pm\sqrt{16}$$ $$x = \pm 4$$

The solutions are $\boxed{x = 4, -4}$.