How do you solve #x^2-5x+3=0#?

1 Answer
Apr 2, 2016

Use the quadratic formula to find roots:

#x = 5/2+-sqrt(13)/2#

Explanation:

This quadratic equation is in the form #ax^2+bx+c = 0# with #a=1#, #b=-5# and #c=3#.

It has roots given by the quadratic formula:

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

#=(5+-sqrt(5^2-(4*1*3)))/(2*1)#

#=(5+-sqrt(25-12))/2#

#=(5+-sqrt(13))/2#

That is #x = 5/2+sqrt(13)/2# or #x = 5/2-sqrt(13)/2#