When you’re solving something like:

$$ax^2 + bx + c = 0 \quad (a \ne 0)$$

what you’re really trying to do is find the values of $x$ that make this equation true.

These values, called the solutions or roots, are the points where the function $f(x) = ax^2 + bx + c$ hits zero.

Visually, that means: where does the parabola cross the x-axis?

Sometimes it touches the axis once. Sometimes twice. Sometimes it floats above or below and never touches at all. But in all cases, there’s still an answer. You just might have to allow complex numbers to get there.

There are a few different ways to solve these equations. Let’s look at how they actually work.

Solving by Factoring

Say you’ve got something like:

$$x^2 – 5x + 6 = 0$$

You’re trying to find values of $x$ that make the whole expression zero. If you can factor it like this:

$$(x – 2)(x – 3) = 0$$

you’ve turned it into a product of two expressions. And if their product is zero, then one of them must be zero. So:

$$x – 2 = 0 \quad \text{or} \quad x – 3 = 0$$

Which gives:

$$x = 2 \quad \text{or} \quad x = 3$$

That’s it. Those are your solutions.

This works cleanly when the quadratic factors nicely, when the numbers are friendly and the patterns are obvious. And to even try factoring, you first need the equation set up like this:

$$a{x}^2+bx+c=0$$

$$ax^2 + bx + c = 0$$

Everything on one side, zero on the other. If you’re missing that setup, you can’t apply the zero-product rule.

Now, if the quadratic doesn’t factor easily, or at all, you’ll want to move to another method. Factoring isn’t a one-size-fits-all approach. It’s just the fastest one when it works.

Completing the Square

This method takes a different approach. Instead of trying to break the quadratic into pieces, you shape it into something that’s easy to solve directly.

The idea is to turn the equation into something like:

$$(x+\text{something}{)}^2=\text{number}$$

$$(x + \text{something})^2 = \text{number}$$

Once it’s in that form, you can just take the square root of both sides and solve.

Let’s walk through it with an example:

$$x^2+6x+5=0$$

$$x^2 + 6x + 5 = 0$$

First, move the constant over:

$$x^2 + 6x = -5$$

Now take half of the 6, square it (you get 9), and add it to both sides:

$$x^2 + 6x + 9 = -5 + 9 \Rightarrow (x + 3)^2 = 4$$

Take the square root:

$$x + 3 = \pm 2$$

Then solve:

$$x = -1 \quad \text{or} \quad x = -5$$

That’s the method. You’re literally completing the expression so it becomes a perfect square, something you can work with more directly.

If the coefficient on $x^2$ isn’t 1, just divide everything by that number first. It adds a step, but the overall process is the same.

This method is clean and systematic. And if it feels a little mechanical, there’s a reason. It’s the foundation for the quadratic formula. That formula you memorized is just a cleaned-up version of completing the square.

The Quadratic Formula

When factoring doesn’t work, and you don’t feel like completing the square every time, the quadratic formula is your all-purpose tool:

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

You plug in the values of $a$, $b$, and $c$, and out come your solutions.

The part under the square root is called the discriminant. It tells you what kind of solutions you’ll get:

If it’s positive, you get two real solutions.
If it’s zero, you get one real solution (what’s called a double root).
If it’s negative, you get two complex solutions involving $i$ (no real solutions).

The formula always works. No guessing, no clever factoring, no tricks. Just numbers in, answers out.

Graphing and Estimating

Sometimes, especially if you’re unsure whether the equation will factor, it helps to see the function.

If you graph $y = ax^2 + bx + c$, the solutions to the equation $ax^2 + bx + c = 0$ are the x-values where the graph touches or crosses the x-axis.

If it doesn’t touch at all, you know the roots are complex. If it just grazes the axis, that’s a double root. And if it cuts through in two places, those x-values are your two real solutions.

If the roots are messy, like irrational numbers that don’t come out nicely, you can estimate them using a calculator or a method like Newton’s iteration. Sometimes a good visual or a close approximation is all you need.

Vieta’s Formulas

If you already know the solutions to a quadratic, there’s a shortcut to check your work or even rebuild the original equation.

Say the roots are $r_1$ and $r_2$. Then:

$$r_1 + r_2 = \frac{-b}{a}, \quad r_1 \cdot r_2 = \frac{c}{a}$$

These relationships come straight from the factored form of the quadratic. You’re just working backwards from:

$$a(x – r_1)(x – r_2)$$

It’s a nice tool to keep in your pocket, especially if you’re checking your answers or building equations from given roots.