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

##### 2 Answers

#### Explanation:

The quadratic formula works for equations of the form:

#y = ax^2 + bx + c#

And states that the solutions for

#x = (-b+-sqrt(b^2-4ac))/(2a)#

In this case, the quadratic equation is

#a = 1#

#b = -5#

#c = 6#

So, just plug these values into the quadratic equation and simplify.

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-(-5)+-sqrt((-5)^2-4(1)(6)))/(2(1))#

#x = (5+-sqrt(25-24))/2#

#x = (5+-sqrt1)/2#

#x = (5-1)/2 color(white)"XX" or color(white)"XX" x = (5+1)/2#

#x = 2 color(white)"XX" or color(white)"XX" x = 3#

Therefore, the solutions to this quadratic are

#x in {2,3}#

This just means that

*Final Answer*

#### Explanation:

The standard form of a quadratic function is.

#• y=ax^2+bx+c ; a!=0#

#"For " y=x^2-5x+6#

#rArra=1,b=-5" and " c=6# Using the

#color(blue)"quadratic formula"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(x=(-b+-sqrt(b^2-4ac))/(2a))color(white)(2/2)|)))#

#rArrx=(-(color(red)(-5))+-sqrt(25-(4xx1xx6)))/2#

#color(white)(rArrx)=(5+-sqrt1)/2#

#rArrx_1=(5+1)/2" or " x_2=(5-1)/2#

#rArr" roots are " x_1=3" or " x_2=2# Both roots are real. There are no imaginary roots.