## Solving the Quadratic Equation: (x^2-x)^2 + 5(x^2-x) + 4 = 0

This equation might look intimidating at first glance, but we can solve it using a clever substitution and the familiar quadratic formula.

### Simplifying the Equation with Substitution

Let's start by making a substitution to simplify the equation. Let **y = x^2 - x**. Now we can rewrite the equation as:

**y^2 + 5y + 4 = 0**

This is a much more manageable quadratic equation.

### Solving the Quadratic Equation

We can now solve for *y* using the quadratic formula:

**y = (-b ± √(b^2 - 4ac)) / 2a**

Where:

- a = 1
- b = 5
- c = 4

Substituting the values into the formula, we get:

**y = (-5 ± √(5^2 - 4 * 1 * 4)) / 2 * 1**

**y = (-5 ± √9) / 2**

This gives us two possible solutions for *y*:

**y1 = -1**

**y2 = -4**

### Solving for x

Now we need to substitute back our original expression for *y* and solve for *x*:

**For y1 = -1:**

- x^2 - x = -1
- x^2 - x + 1 = 0

This quadratic equation doesn't factor easily. We can use the quadratic formula again to find the solutions for *x*:

**x = (1 ± √(1^2 - 4 * 1 * 1)) / 2 * 1**

**x = (1 ± √(-3)) / 2**

**x = (1 ± i√3) / 2**

Where *i* is the imaginary unit (√-1). This gives us two complex solutions for *x*.

**For y2 = -4:**

- x^2 - x = -4
- x^2 - x + 4 = 0

Again, this equation doesn't factor easily. Using the quadratic formula:

**x = (1 ± √(1^2 - 4 * 1 * 4)) / 2 * 1**

**x = (1 ± √(-15)) / 2**

**x = (1 ± i√15) / 2**

This also gives us two complex solutions for *x*.

### Conclusion

In conclusion, the equation (x^2-x)^2 + 5(x^2-x) + 4 = 0 has **four complex solutions:**

**x = (1 + i√3) / 2****x = (1 - i√3) / 2****x = (1 + i√15) / 2****x = (1 - i√15) / 2**