# How do you find the zeros, real and imaginary, of y= x^2-5x+6 using the quadratic formula?

Jun 18, 2018

$x = 3 ,$ $2$

#### Explanation:

$y = {x}^{2} - 5 x + 6$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 1$, $b = - 5$, $c = 6$

The zeroes are the values for $x$ when $y = 0$.

Substitute $0$ for $y$.

$0 = {x}^{2} - 5 x + 6$

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

Plug in the known values and solve.

$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

$x = \frac{5 \pm \sqrt{1}}{2}$

Simplify.

$x = \frac{5 \pm 1}{2}$

$x = \frac{5 + 1}{2}$, $\frac{5 - 1}{2}$

$x = \frac{6}{2}$, $\frac{4}{2}$

Simplify.

$x = 3 ,$ $2$

graph{y=x^2-5x+6 [-10, 10, -5, 5]}