Let's solve these problems step by step:

01. $x^2 - 5x = -6$

Rewriting the equation in standard quadratic form:
$x^2 - 5x + 6 = 0$
Factorize the quadratic:
$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$
Using the zero-product property:
$x - 2 = 0 \quad \text{or} \quad x - 3 = 0$
Thus:
$x = 2 \quad \text{or} \quad x = 3$

02. $x^2 - x - 6 = 0$

This is already in standard quadratic form. Factorize:
$x^2 - x - 6 = (x - 3)(x + 2) = 0$
Using the zero-product property:
$x - 3 = 0 \quad \text{or} \quad x + 2 = 0$
Thus:
$x = 3 \quad \text{or} \quad x = -2$

03. $(x + 2)(x - 5) = 0$

Using the zero-product property directly:
$x + 2 = 0 \quad \text{or} \quad x - 5 = 0$
Thus:
$x = -2 \quad \text{or} \quad x = 5$

04. Rectangular Plot Problem

We are given:

• Length = $2x + 1$
• Breadth = $x$
• Area = $36 \, \text{m}^2$

From the area formula ($\text{Area} = \text{Length} \times \text{Breadth}$):
$(2x + 1) \cdot x = 36$
Simplify:
$2x^2 + x = 36$
Rearrange into standard quadratic form:
$2x^2 + x - 36 = 0$
Solve this quadratic using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 1$, $c = -36$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-36)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 288}}{4}$
$x = \frac{-1 \pm \sqrt{289}}{4}$
$x = \frac{-1 \pm 17}{4}$
Solve for both cases:

1. $x = \frac{-1 + 17}{4} = \frac{16}{4} = 4$
2. $x = \frac{-1 - 17}{4} = \frac{-18}{4} = -4.5$

Since $x$ must be positive (it represents length):
$x = 4$

Now, calculate the length ($2x + 1$) and breadth ($x$):

• Length: $2(4) + 1 = 9 \, \text{m}$
• Breadth: $x = 4 \, \text{m}$

The perimeter of the rectangle is:
$P = 2(\text{Length} + \text{Breadth}) = 2(9 + 4) = 2(13) = 26 \, \text{m}$

If three strands of wire are required, the total length of wire is:
$3 \times 26 = 78 \, \text{m}$

05. $2x^2 + 5x - 7 = 0$

Solve using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, $a = 2$, $b = 5$, $c = -7$:
$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-7)}}{2(2)}$
$x = \frac{-5 \pm \sqrt{25 + 56}}{4}$
$x = \frac{-5 \pm \sqrt{81}}{4}$
$x = \frac{-5 \pm 9}{4}$

Solve for both cases:

1. $x = \frac{-5 + 9}{4} = \frac{4}{4} = 1$
2. $x = \frac{-5 - 9}{4} = \frac{-14}{4} = -3.5$

Thus:
$x = 1 \quad \text{or} \quad x = -3.5$

Summary of Solutions:

1. $x = 2, 3$
2. $x = 3, -2$
3. $x = -2, 5$
4. Total wire length = $78 \, \text{m}$
5. $x = 1, -3.5$

Do you want further explanation or clarification on any part?

Related Questions:

1. How do you derive the quadratic formula?
2. What is the discriminant, and how does it determine the nature of roots?
3. How do you calculate the perimeter of any polygon?
4. Can you explain why negative lengths are not physically valid in geometry?
5. What are the steps to factorize a quadratic equation?

Tip: Always rewrite equations into standard quadratic form ($ax^2 + bx + c = 0$) before attempting to solve them!