# How do you solve y=x^2-12x+36?

Jul 5, 2015

This is a perfect square trinomial:

${x}^{2} - 12 x + 36 = {\left(x - 6\right)}^{2}$

So $y = 0$ when $x = 6$

#### Explanation:

Perfect square trinomials are of the form:

${a}^{2} \pm 2 a b + {b}^{2} = {\left(a \pm b\right)}^{2}$

In our case notice that the leading term ${x}^{2}$ is square and the constant term $36 = {6}^{2}$ is square. So the question is: Is the middle term equal to $\pm 2 \cdot x \cdot 6 = \pm 12 x$ - Yes.