We need to solve the following equations: (a) $x^2 - 2x - 8 = 0$ (b) $y^2 = 3y + 4$ (c) $2p^2 - 7p - 4 = 0$ (d) $4m^2 = -3(4m+3)$ (e) $\frac{2q-7}{3} = \frac{q+6}{3q-4}$ (f) $\frac{p+1}{5} = \frac{7-p}{6p}$

1. Problem Description

We need to solve the following equations:

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

(b) $y^2 = 3y + 4$

(c) $2p^2 - 7p - 4 = 0$

(d) $4m^2 = -3(4m+3)$

(e) $\frac{2q-7}{3} = \frac{q+6}{3q-4}$

(f) $\frac{p+1}{5} = \frac{7-p}{6p}$

2. Solution Steps

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

We can factor this quadratic equation:

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

Thus, $x = 4$ or $x = -2$

(b) $y^2 = 3y + 4$

$y^2 - 3y - 4 = 0$

$(y-4)(y+1) = 0$

Thus, $y = 4$ or $y = -1$

(c) $2p^2 - 7p - 4 = 0$

We can factor this quadratic equation:

$(2p+1)(p-4) = 0$

Thus, $2p+1 = 0$ or $p-4 = 0$

$p = -\frac{1}{2}$ or $p = 4$

(d) $4m^2 = -3(4m+3)$

$4m^2 = -12m - 9$

$4m^2 + 12m + 9 = 0$

$(2m+3)(2m+3) = 0$

$(2m+3)^2 = 0$

$2m+3 = 0$

$m = -\frac{3}{2}$

(e) $\frac{2q-7}{3} = \frac{q+6}{3q-4}$

$(2q-7)(3q-4) = 3(q+6)$

$6q^2 - 8q - 21q + 28 = 3q + 18$

$6q^2 - 29q + 28 = 3q + 18$

$6q^2 - 32q + 10 = 0$

$3q^2 - 16q + 5 = 0$

$(3q-1)(q-5) = 0$

Thus, $3q-1 = 0$ or $q-5 = 0$

$q = \frac{1}{3}$ or $q = 5$

(f) $\frac{p+1}{5} = \frac{7-p}{6p}$

$6p(p+1) = 5(7-p)$

$6p^2 + 6p = 35 - 5p$

$6p^2 + 11p - 35 = 0$

$(6p + 3)(p-\frac{35}{3})$

$(2p+5)(3p-7) = 0$

Thus, $2p+5=0$ or $3p-7=0$

$p = -\frac{5}{2}$ or $p = \frac{7}{3}$

3. Final Answer

(a) $x = 4, -2$

(b) $y = 4, -1$

(c) $p = -\frac{1}{2}, 4$

(d) $m = -\frac{3}{2}$

(e) $q = \frac{1}{3}, 5$

(f) $p = -\frac{5}{2}, \frac{7}{3}$