Let's go through each of the equations one by one.

Question 13:

$x^2 = -x + 12$ To solve, move all terms to one side of the equation: $x^2 + x - 12 = 0$ Now, solve this quadratic equation by factoring, completing the square, or using the quadratic formula. We'll try factoring first: $x^2 + x - 12 = (x + 4)(x - 3) = 0$ So, the solutions are: $x = -4 \quad \text{or} \quad x = 3$

Question 14:

$42 = x^2 + x$ Move all terms to one side: $x^2 + x - 42 = 0$ Now, factor the quadratic: $x^2 + x - 42 = (x + 7)(x - 6) = 0$ So, the solutions are: $x = -7 \quad \text{or} \quad x = 6$

Question 15:

$10x^2 - 48 = -74x$ Move all terms to one side: $10x^2 + 74x - 48 = 0$ This is a quadratic equation. Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For $a = 10$, $b = 74$, and $c = -48$: $x = \frac{-74 \pm \sqrt{74^2 - 4(10)(-48)}}{2(10)}$ $x = \frac{-74 \pm \sqrt{5476 + 1920}}{20}$ $x = \frac{-74 \pm \sqrt{7396}}{20}$ $x = \frac{-74 \pm 86}{20}$ So, the two solutions are: $x = \frac{-74 + 86}{20} = \frac{12}{20} = 0.6$ $x = \frac{-74 - 86}{20} = \frac{-160}{20} = -8$

Question 16:

$x^2 + x = 2$ Move all terms to one side: $x^2 + x - 2 = 0$ Now, factor the quadratic: $x^2 + x - 2 = (x - 1)(x + 2) = 0$ So, the solutions are: $x = 1 \quad \text{or} \quad x = -2$

Summary of solutions:

1. $x = -4$ or $x = 3$ for Question 13.
2. $x = -7$ or $x = 6$ for Question 14.
3. $x = 0.6$ or $x = -8$ for Question 15.
4. $x = 1$ or $x = -2$ for Question 16.

Would you like further explanations or any of these solutions broken down in more detail?

Here are 5 related questions you might find useful:

1. How do you complete the square for solving quadratic equations?
2. Can all quadratic equations be factored easily?
3. When should you use the quadratic formula instead of factoring?
4. What is the discriminant in a quadratic equation, and what does it tell you?
5. How do you solve quadratics with coefficients that aren't simple integers?

Tip: When factoring quadratics, always look for patterns or factorable pairs that multiply to give the constant term and add to give the coefficient of $x$.