# How do you find the zeros, real and imaginary, of y=x^2-x+7 using the quadratic formula?

May 3, 2017

Plug in the given values for $a$, $b$, and $c$ into the formula, and solve.
$x = \frac{1 \pm 3 i \sqrt{3}}{2}$.

#### Explanation:

Finding the zeroes (or roots) of a quadratic equation means solving for $x$ when $y = 0$. In other words, we want to know what values of $x$ we would have to plug into ${x}^{2} - x + 7$ in order for it to simplify to $0.$

To do this, there are some shortcuts that might be easy to see, but the quadratic formula is always an option. This formula states that, if we want $a {x}^{2} + b x + c$ to equal $0 ,$ then the $x$-values that make this true will be

x=(–b+-sqrt(b^2-4ac))/(2a)

So, for any specific quadratic expression (like ${x}^{2} - x + 7$), all we need to do is plug the given values for $a , b ,$ and $c$ into the quadratic formula.

For ${x}^{2} - x + 7$, we have

$a = 1$ (from the $1 {x}^{2}$),
b=–1 (from the $- 1 x$), and
$c = 7$ (from the $+ 7$).

Plugging these into the quadratic formula gives

x=(–color(purple)b+-sqrt(color(blue)b^2-4color(orange)acolor(green)c))/(2color(orange)a)

x=(–color(blue)((–1))+-sqrt(color(blue)((–1))^2-4color(orange)((1))color(green)((7))))/(2color(orange)((1)))

$x = \frac{1 \pm \sqrt{1 - 28}}{2}$

x=(1+-sqrt(–27))/(2)

x=(1+-3sqrt(–3))/(2)

Because we get a negative under the square root, the zeroes are imaginary. Recalling that i^2=–1, we get

$x = \frac{1 \pm 3 i \sqrt{3}}{2}$.