# How do you find the zeros, if any, of y=x^2-3x+1 using the quadratic formula?

May 6, 2018

The solutions are $\setminus \frac{3 \pm \sqrt{5}}{2}$

#### Explanation:

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

the quadratic formula states that the solutions, if any, are

${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

The quantity $\setminus \Delta = {b}^{2} - 4 a c$ is called Determinant, and you can tell the number of solutions by its sign:

1. If $\Delta > 0$, then the square root is well defined and not null, so you have two solutions.
2. If $\Delta = 0$, the square root is zero as well. So, adding or subtracting it makes no difference, and you have two coincident solutions.
3. If $\Delta < 0$, the square root is not defined, and you have no solutions (at least using real numbers).

In your case, $a = 1$,$b = - 3$ and $c = 1$. So, you have

$\Delta = {\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 1 = 9 - 4 = 5$, so you have two solutions, namely

${x}_{1 , 2} = \setminus \frac{- \left(- 3\right) \setminus \pm \setminus \sqrt{5}}{2 \cdot 1} = \setminus \frac{3 \pm \sqrt{5}}{2}$