The equation you provided is:

$x^2 + x = 4$

To solve for $x$, follow these steps:

1. Rearrange the equation to set it equal to zero: $x^2 + x - 4 = 0$

   This is a quadratic equation of the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 1$, and $c = -4$.

2. Use the quadratic formula to solve for $x$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting the values of $a$, $b$, and $c$ into the formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)}$ Simplify: $x = \frac{-1 \pm \sqrt{1 + 16}}{2}$ $x = \frac{-1 \pm \sqrt{17}}{2}$ Therefore, the solutions are: $x = \frac{-1 + \sqrt{17}}{2} \quad \text{or} \quad x = \frac{-1 - \sqrt{17}}{2}$

These are the two possible solutions for $x$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you complete the square to solve quadratic equations?
2. What are the steps to solve a quadratic equation using factoring?
3. How is the discriminant in the quadratic formula interpreted?
4. How do you graph the equation $x^2 + x - 4 = 0$?
5. What real-life situations can quadratic equations model?

Tip: The discriminant $b^2 - 4ac$ in the quadratic formula tells you how many real solutions the equation has: if it's positive, there are two real solutions.