The problem asks us to find the solutions of several equations. The equations are: a) $2x + 3 = -3 + 3x$ b) $\frac{2}{3}x + \frac{5}{4}$ c) $(2x - 3)(4 + 2x) = 0$ d) $x^2 - 5x$ e) $x^2 + 2x + 1 = 0$ f) $x^2 + 4$

1. Problem Description

The problem asks us to find the solutions of several equations. The equations are:

a) $2x + 3 = -3 + 3x$

b) $\frac{2}{3}x + \frac{5}{4}$

c) $(2x - 3)(4 + 2x) = 0$

d) $x^2 - 5x$

e) $x^2 + 2x + 1 = 0$

f) $x^2 + 4$

2. Solution Steps

a) $2x + 3 = -3 + 3x$

Subtract $2x$ from both sides:

$3 = -3 + x$

Add 3 to both sides:

$6 = x$

$x = 6$

b) This equation is incomplete. It needs to be set equal to something. Without knowing what it equals, we cannot solve the equation. We assume it should have been $= 0$. So the question becomes $\frac{2}{3}x + \frac{5}{4} = 0$

Subtract $\frac{5}{4}$ from both sides:

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

Multiply both sides by $\frac{3}{2}$:

$x = -\frac{5}{4} \cdot \frac{3}{2}$

$x = -\frac{15}{8}$

c) $(2x - 3)(4 + 2x) = 0$

If the product of two factors is zero, then at least one of the factors must be zero.

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

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

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

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

Factor out $x$:

$x(x - 5) = 0$

$x = 0$ or $x - 5 = 0$

$x = 0$ or $x = 5$

e) $x^2 + 2x + 1 = 0$

This is a perfect square trinomial:

$(x + 1)^2 = 0$

$x + 1 = 0$

$x = -1$

f) $x^2 + 4 = 0$

$x^2 = -4$

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

$x = \pm 2i$

3. Final Answer

a) $x = 6$

b) $x = -\frac{15}{8}$

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

d) $x = 0, 5$

e) $x = -1$

f) $x = 2i, -2i$