To solve the given equations by transforming them into standard form $ax^2 + bx + c = 0$ and solving for $x$ using factoring or the quadratic formula, we will follow the steps outlined for each equation.

1. $x(x+3) = 4$

First, expand and bring the equation to standard form: $x(x+3) = 4$ $x^2 + 3x = 4$ Subtract 4 from both sides: $x^2 + 3x - 4 = 0$ Now, solve this quadratic equation by factoring: $(x + 4)(x - 1) = 0$ The solutions are: $x + 4 = 0 \implies x = -4$ $x - 1 = 0 \implies x = 1$ So, the solutions are $x = -4$ and $x = 1$.

2. $x(x+3) = 10$

Expand and bring the equation to standard form: $x(x+3) = 10$ $x^2 + 3x = 10$ Subtract 10 from both sides: $x^2 + 3x - 10 = 0$ Now, solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 3$, and $c = -10$: $x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)}$ $x = \frac{-3 \pm \sqrt{9 + 40}}{2}$ $x = \frac{-3 \pm \sqrt{49}}{2}$ $x = \frac{-3 \pm 7}{2}$ So, the solutions are: $x = \frac{-3 + 7}{2} = 2 \quad \text{and} \quad x = \frac{-3 - 7}{2} = -5$ Thus, the solutions are $x = 2$ and $x = -5$.

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

Expand both sides: $x(2x+1) = 2(x-1)$ $2x^2 + x = 2x - 2$ Move all terms to one side of the equation: $2x^2 + x - 2x + 2 = 0$ Simplify: $2x^2 - x + 2 = 0$ Use the quadratic formula to solve: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(2)}}{2(2)}$ $x = \frac{1 \pm \sqrt{1 - 16}}{4}$ $x = \frac{1 \pm \sqrt{-15}}{4}$ Since the discriminant is negative, the solutions are complex: $x = \frac{1 \pm i\sqrt{15}}{4}$ So, the solutions are $x = \frac{1 + i\sqrt{15}}{4}$ and $x = \frac{1 - i\sqrt{15}}{4}$.

4. $(x + 5)^2 + (x - 2)^2 = 37$

Expand both terms: $(x + 5)^2 + (x - 2)^2 = 37$ $(x^2 + 10x + 25) + (x^2 - 4x + 4) = 37$ Combine like terms: $2x^2 + 6x + 29 = 37$ Subtract 37 from both sides: $2x^2 + 6x - 8 = 0$ Simplify by dividing the equation by 2: $x^2 + 3x - 4 = 0$ Factor the quadratic equation: $(x + 4)(x - 1) = 0$ The solutions are: $x + 4 = 0 \implies x = -4$ $x - 1 = 0 \implies x = 1$ Thus, the solutions are $x = -4$ and $x = 1$.

5. $(x + 6) - 15 = 0$

Simplify: $x + 6 - 15 = 0$ $x - 9 = 0$ Solve for $x$: $x = 9$

---

Let me know if you want more details on any of the steps! Here are five related questions to explore further:

1. How do you use the discriminant to determine the nature of roots in a quadratic equation?
2. Can all quadratic equations be factored easily, or are some better suited for the quadratic formula?
3. What is the significance of completing the square, and how does it help solve quadratic equations?
4. How can graphing help visualize the solutions to quadratic equations?
5. How does the quadratic formula relate to the general form of a quadratic equation?

Tip: When factoring, always check for the greatest common factor (GCF) first to simplify the equation.