How to Solve x^2 - 5x + 6 = 0 Step by Step

A quadratic equation is any equation where the highest power of x is 2. The general shape is ax^2 + bx + c = 0, with a not equal to zero. In our puzzle, a = 1, b = -5 and c = 6.

Why does the power matter? Because a squared term makes the graph a U-shaped curve called a parabola. That curve can cross the x-axis at two points, one point, or not at all. Those crossing points are the solutions we're hunting for.

So when we say solve x^2 - 5x + 6 = 0, we're really asking: at which x values does this parabola touch the x-axis? For this equation, the answer is two points, x = 2 and x = 3.

Factoring is reverse multiplication. When you expand (x - 2)(x - 3), you multiply each term and get back x^2 - 5x + 6. So writing the equation in factored form doesn't change it, it just shows its hidden structure.

The magic step is the zero product rule. If two things multiply to give zero, at least one of them must be zero. Think about it: 5 times 0 is 0, and 0 times anything is 0. There's no other way to land on zero by multiplying.

That rule is what lets us split one equation into two simple ones. It's the heart of the whole method, so it's worth saying out loud: if (x - 2)(x - 3) = 0, then x - 2 = 0 or x - 3 = 0.

Here's the full process, one small step at a time. Follow along with a pencil, it sticks better that way.

• Write the equation in standard form: x^2 - 5x + 6 = 0. It already is, good.
• Find two numbers that multiply to +6 (the constant term) and add up to -5 (the coefficient of x).
• Those numbers are -2 and -3, because (-2) x (-3) = 6 and (-2) + (-3) = -5.
• Rewrite the middle term using them: x^2 - 2x - 3x + 6 = 0.
• Group and factor: x(x - 2) - 3(x - 2) = 0, which gives (x - 2)(x - 3) = 0.
• Set each factor to zero: x - 2 = 0 or x - 3 = 0.
• Solve each: x = 2 or x = 3. Done!

Plot y = x^2 - 5x + 6 and you get a U-shaped parabola opening upwards. It cuts the x-axis exactly at x = 2 and x = 3. Those crossing points are the solutions, also called roots or zeros.

This gives you a nice picture in your head. Solving the equation and finding where the curve meets the x-axis are the same job, just described in two languages, algebra and geometry.

If the parabola sat on the axis at only one point, you'd get one repeated solution. If it floated above the axis without touching, there would be no real solutions. Here we're lucky: two clean crossings.

Yes, and it's good to know a backup. The quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, works on every quadratic, whether it factors neatly or not.

For our equation, b^2 - 4ac = 25 - 24 = 1. So x = (5 ± 1) / 2, which gives x = 3 or x = 2. Same answers, different road.

Factoring is faster when the numbers are friendly, like here. The formula is your reliable friend when the numbers get messy. Knowing both means you're never stuck.

Try these in your head or on paper. Each one uses the same idea: find two numbers that multiply to the constant and add to the middle coefficient.

• x^2 - 7x + 12 = 0: the numbers are -3 and -4, so x = 3 or x = 4.
• x^2 + 5x + 6 = 0: the numbers are +2 and +3, so x = -2 or x = -3. Notice the sign flip.
• x^2 - x - 6 = 0: the numbers are -3 and +2, so x = 3 or x = -2.
• If you get stuck, fall back on the quadratic formula. It never lets you down.