 using factoring or Quadratic formula.
1. x(x+3) =4
2. x(x+3) =10
3. x(2x+1)= 2(x-1)
4. (x + 5)²+ (x- 2)²=37
5. (x + 6)-15

Let's solve each problem step-by-step using either factoring or the quadratic formula.

### 1. Solving \(x(x + 3) = 4\)

**Step 1: Expand the equation:**

\[
x^2 + 3x = 4
\]

**Step 2: Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):**

\[
x^2 + 3x - 4 = 0
\]

**Step 3: Factor the quadratic expression:**

We look for two numbers that multiply to \(-4\) and add up to \(3\). The numbers are \(4\) and \(-1\):

\[
x^2 + 4x - x - 4 = 0
\]

\[
(x - 1)(x + 4) = 0
\]

**Step 4: Solve for \(x\):**

\[
x - 1 = 0 \implies x = 1
\]
\[
x + 4 = 0 \implies x = -4
\]

**Solutions:** \(x = 1\) and \(x = -4\)

### 2. Solving \(x(x + 3) = 10\)

**Step 1: Expand the equation:**

\[
x^2 + 3x = 10
\]

**Step 2: Rearrange into standard quadratic form:**

\[
x^2 + 3x - 10 = 0
\]

**Step 3: Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), and \(c = -10\):**

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)}
\]

\[
x = \frac{-3 \pm \sqrt{9 + 40}}{2}
\]

\[
x = \frac{-3 \pm \sqrt{49}}{2}
\]

\[
x = \frac{-3 \pm 7}{2}
\]

**Step 4: Solve for \(x\):**

\[
x_1 = \frac{-3 + 7}{2} = \frac{4}{2} = 2
\]

\[
x_2 = \frac{-3 - 7}{2} = \frac{-10}{2} = -5
\]

**Solutions:** \(x = 2\) and \(x = -5\)

### 3. Solving \(x(2x + 1) = 2(x - 1)\)

**Step 1: Expand both sides:**

\[
2x^2 + x = 2x - 2
\]

**Step 2: Rearrange into standard quadratic form:**

\[
2x^2 + x - 2x + 2 = 0
\]

\[
2x^2 - x + 2 = 0
\]

**Step 3: Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -1\), and \(c = 2\):**

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(2)}}{2(2)}
\]

\[
x = \frac{1 \pm \sqrt{1 - 16}}{4}
\]

\[
x = \frac{1 \pm \sqrt{-15}}{4}
\]

**Step 4: Solve for \(x\):**

Since the discriminant is negative, there are no real solutions.

**Solutions:** No real solutions.

### 4. Solving \((x + 5)^2 + (x - 2)^2 = 37\)

**Step 1: Expand both squared terms:**

\[
(x + 5)^2 = x^2 + 10x + 25
\]
\[
(x - 2)^2 = x^2 - 4x + 4
\]

**Step 2: Combine them and simplify:**

\[
x^2 + 10x + 25 + x^2 - 4x + 4 = 37
\]

\[
2x^2 + 6x + 29 = 37
\]

**Step 3: Rearrange into standard quadratic form:**

\[
2x^2 + 6x - 8 = 0
\]

**Step 4: Divide by 2 to simplify:**

\[
x^2 + 3x - 4 = 0
\]

**Step 5: Factor the quadratic expression:**

\[
(x - 1)(x + 4) = 0
\]

**Step 6: Solve for \(x\):**

\[
x - 1 = 0 \implies x = 1
\]
\[
x + 4 = 0 \implies x = -4
\]

**Solutions:** \(x = 1\) and \(x = -4\)

### 5. Solving \((x + 6) - 15 = 0\)

**Step 1: Simplify the expression:**

\[
x + 6 - 15 = 0
\]

\[
x - 9 = 0
\]

**Step 2: Solve for \(x\):**

\[
x = 9
\]

**Solution:** \(x = 9\)

---

Would you like more details on any of these steps, or do you have any questions?

### Related Questions:
1. How do you solve a quadratic equation by completing the square?
2. What is the quadratic formula and when is it used?
3. How do you determine the number of real solutions for a quadratic equation?
4. Can all quadratic equations be factored? Why or why not?
5. What is the significance of the discriminant in a quadratic equation?

**Tip:** When solving quadratic equations, always check whether the discriminant (\(b^2 - 4ac\)) is positive, zero, or negative. This will tell you if the solutions are real or complex.

Using factoring or Quadratic formula.
1. x(x+3) = 4
2. x(x+3) = 10
3. x(2x+1) = 2(x-1)
4. (x + 5)²+ (x- 2)²=37
5. (x + 6)-15

Learn how to solve quadratic equations such as x(x+3)=4, x(2x+1)=2(x-1), and others using both factoring and the quadratic formula. This tutorial provides step-by-step explanations suitable for students in Grades 8-10.

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