Do I always have to set the equation to zero?

Yes, for factoring and the quadratic formula, one side of the equation must be exactly zero. The only time you do not need a zero is when you are using the square root method.

Will the quadratic formula always work?

Yes, absolutely always. Even if an equation can be factored, the quadratic formula will still give you the correct answer. It just might take a little longer.

What about completing the square?

Completing the square is another method that always works, but it often introduces messy fractions if $a$ is not $1$ or $b$ is an odd number. It is usually best to save it for when a teacher specifically asks for it, or when you are trying to find the vertex of a parabola.

How do you know which method to use to solve a quadratic equation?

Choosing a method to solve a quadratic equation is like picking a tool from a toolbox. If you have a loose screw, you grab a screwdriver; if you have a nail, you grab a hammer. In algebra, you look at the structure of the equation $ax^2 + bx + c = 0$ to pick the best tool.

If the equation is missing the middle $x$ term, you can just take a square root. If the numbers are small and easy to work with, factoring is usually the fastest route. If the numbers are messy or you are stuck, the quadratic formula is your heavy-duty hammer that works every single time.

Taking Square Roots

Use this method when your equation is missing the $x$ term (meaning $b=0$ ), like $x^2 - 16 = 0$ , or when you have a perfect square isolated, like $(x-3)^2 = 25$ . This is the quickest method because you just isolate the squared part and take the square root of both sides. Just remember that taking a square root gives you two answers: a positive and a negative one.

Factoring

Factoring is your go-to method when the numbers play nice. You use this when you can easily think of two numbers that multiply to $c$ and add to $b$ . For example, $x^2 + 5x + 6 = 0$ factors cleanly into $(x+2)(x+3) = 0$ . If you stare at an equation for more than 30 seconds and cannot find the factors, drop this method and move on to the quadratic formula.

The Quadratic Formula

The quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , is your universal fallback. Use this when the equation does not factor easily, when there are decimals or fractions, or when the solutions are imaginary numbers. It is a bit more work, but it is guaranteed to solve any quadratic equation in the universe.

Where students slip up

The biggest mistake students make is not setting the equation to zero before trying to factor or use the quadratic formula. If you have $x^2 + 4x = 5$ , you must move the $5$ over to get $x^2 + 4x - 5 = 0$ first. Another common slip is forgetting the $\pm$ symbol when using the square root method, which causes you to lose half of your answers.

Worked through

Solve the equation $2x^2 - 5x = 3$ by choosing the most appropriate method.

First, we need to set the equation to zero. Subtract 3 from both sides to get $2x^2 - 5x - 3 = 0$ .

Next, we check our methods. There is an $x$ term, so we cannot just take square roots. Let us try factoring. We need two numbers that multiply to $(2)(-3) = -6$ and add to $-5$ . Those numbers are $-6$ and $1$ . Because we found factors quickly, factoring is a great method here.

We rewrite the middle term: $2x^2 - 6x + 1x - 3 = 0$ .

Factor by grouping: $2x(x - 3) + 1(x - 3) = 0$ , which gives $(2x + 1)(x - 3) = 0$ .

Setting each factor to zero gives $2x + 1 = 0$ (so $x = -1/2$ ) and $x - 3 = 0$ (so $x = 3$ ). The solutions are $x = -1/2$ and $x = 3$ .

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Where this comes from: OpenStax Algebra and Trigonometry, Chapter 2: Equations and Inequalities · Khan Academy: Quadratic equations unit