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

##### 1 Answer

#### Answer:

Plug in the given values for

#### Explanation:

Finding the zeroes (or roots) of a quadratic equation means solving for

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

#x=(–b+-sqrt(b^2-4ac))/(2a)#

So, for any specific quadratic expression (like **plug the given values for #a, b,# and #c# into the quadratic formula.**

For

#a=1# (from the#1x^2# ),

#b=–1# (from the#-1x# ), 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=(1+-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

#x=(1+-3isqrt(3))/(2)# .