The problem asks us to find the solution set for the given quadratic equations by factoring.

1. Problem Description

The problem asks us to find the solution set for the given quadratic equations by factoring.

2. Solution Steps

Here are the solutions for each equation:

a) $x^2 + 9x + 14 = 0$

$(x+2)(x+7) = 0$

$x+2 = 0$ or $x+7 = 0$

$x = -2$ or $x = -7$

b) $x^2 - 4x = 12$

$x^2 - 4x - 12 = 0$

$(x-6)(x+2) = 0$

$x-6 = 0$ or $x+2 = 0$

$x = 6$ or $x = -2$

c) $x^2 = 49$

$x^2 - 49 = 0$

$(x-7)(x+7) = 0$

$x-7 = 0$ or $x+7 = 0$

$x = 7$ or $x = -7$

d) $x^2 - 81 = 0$

$(x-9)(x+9) = 0$

$x-9 = 0$ or $x+9 = 0$

$x = 9$ or $x = -9$

e) $9x^2 - 100 = 0$

$(3x-10)(3x+10) = 0$

$3x-10 = 0$ or $3x+10 = 0$

$3x = 10$ or $3x = -10$

$x = \frac{10}{3}$ or $x = -\frac{10}{3}$

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

$4x^2 = 1$

$x^2 = \frac{1}{4}$

$(2x-1)(2x+1) = 0$

$x = \frac{1}{2}$ or $x = -\frac{1}{2}$

g) $3x^2 - 24x = 48$

$3x^2 - 24x - 48 = 0$

$x^2 - 8x - 16 = 0$

This does not factor easily. Let's use the quadratic formula:

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

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-16)}}{2(1)}$

$x = \frac{8 \pm \sqrt{64 + 64}}{2}$

$x = \frac{8 \pm \sqrt{128}}{2}$

$x = \frac{8 \pm 8\sqrt{2}}{2}$

$x = 4 \pm 4\sqrt{2}$

h) $10x^2 = -19x + 15$

$10x^2 + 19x - 15 = 0$

$(2x+5)(5x-3) = 0$

$2x+5 = 0$ or $5x-3 = 0$

$2x = -5$ or $5x = 3$

$x = -\frac{5}{2}$ or $x = \frac{3}{5}$

i) $-8x^2 + 26x - 15 = 0$

$8x^2 - 26x + 15 = 0$

$(4x-3)(2x-5) = 0$

$4x-3 = 0$ or $2x-5 = 0$

$4x = 3$ or $2x = 5$

$x = \frac{3}{4}$ or $x = \frac{5}{2}$

j) $2x^2 - 4x = 16$

$2x^2 - 4x - 16 = 0$

$x^2 - 2x - 8 = 0$

$(x-4)(x+2) = 0$

$x-4 = 0$ or $x+2 = 0$

$x = 4$ or $x = -2$

k) $3x^2 = 19x - 20$

$3x^2 - 19x + 20 = 0$

$(3x-4)(x-5) = 0$

$3x-4 = 0$ or $x-5 = 0$

$3x = 4$ or $x = 5$

$x = \frac{4}{3}$ or $x = 5$

l) $6x^2 + 11x = 10$

$6x^2 + 11x - 10 = 0$

$(2x+5)(3x-2) = 0$

$2x+5 = 0$ or $3x-2 = 0$

$2x = -5$ or $3x = 2$

$x = -\frac{5}{2}$ or $x = \frac{2}{3}$

m) $(x+2)^2 = 4$

$(x+2)^2 - 4 = 0$

$(x+2-2)(x+2+2) = 0$

$x(x+4) = 0$

$x = 0$ or $x+4 = 0$

$x = 0$ or $x = -4$

n) $(x-5)^2 - 9 = 0$

$((x-5)-3)((x-5)+3) = 0$

$(x-8)(x-2) = 0$

$x-8 = 0$ or $x-2 = 0$

$x = 8$ or $x = 2$

ñ) $6x^2 - 15x - 9 = 0$

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

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

$2x+1 = 0$ or $x-3 = 0$

$2x = -1$ or $x = 3$

$x = -\frac{1}{2}$ or $x = 3$

o) $x^2 - 1 = 0$

$(x-1)(x+1) = 0$

$x-1 = 0$ or $x+1 = 0$

$x = 1$ or $x = -1$

3. Final Answer

a) $x = -2, -7$

b) $x = 6, -2$

c) $x = 7, -7$

d) $x = 9, -9$

e) $x = \frac{10}{3}, -\frac{10}{3}$

f) $x = \frac{1}{2}, -\frac{1}{2}$

g) $x = 4 + 4\sqrt{2}, 4 - 4\sqrt{2}$

h) $x = -\frac{5}{2}, \frac{3}{5}$

i) $x = \frac{3}{4}, \frac{5}{2}$

j) $x = 4, -2$

k) $x = \frac{4}{3}, 5$

l) $x = -\frac{5}{2}, \frac{2}{3}$

m) $x = 0, -4$

n) $x = 8, 2$

ñ) $x = -\frac{1}{2}, 3$

o) $x = 1, -1$