It's a quadratic equation and the x^2 is x squared

## How to solve this quadratic equation x^2-4x-5=0

# 2 Answers

The goal here is to solve for the unknown variable, **x**, when the equation is equal to zero. Since the equation is a 2nd degree polynomial function (or a quadratic function), we first want to change it into a product of two binomials by factoring.
Once we have done that, we can apply the zero product property to solve for x.

x^{2} - 4x - 5 = 0

(x - 5)(x + 1) = 0

x - 5 = 0 and x + 1 = 0

**x = 5** **x = -1**

There are two ways you could do this: factoring and then using the zero product rule, or using the quadratic formula.

The Quadratic Formula

x = [-b ± √(b^{2} - 4ac)]/2a, where your original quadratic equation takes the form ax^{2} + bx + c = 0

In your problem, a = 1, b = -4, and c = -5

When you plug these values into the formula and solve for x, you will get two different answers -- one for adding in the numerator, and one for subtracting in the numerator.

Zero Product Rule

When you have (x+a)(x+b)=0, your two solutions will be **a** and **b**.

One of these must be true in order for the product to equal 0 (Zero Product Rule): x+a=0, OR x+b=0

To factor your equation to make it look like that, find two things that multiply to equal -5, and add up to -4: -5 and 1.

Now factor your equation: x^{2} - 4x - 5 = 0 ===> (x-5)(x+1) = 0

x-5=0 ===> **x = 5**

x+1=0 ===> **x = -1**