How do you solve #2x^2=6x-2# using the quadratic formula?

1 Answer
Feb 20, 2016

Answer:

#x=(3+sqrt(5))/2# or #(3-sqrt(5))/2#

Explanation:

Using quadratic formula, solution of the equation #ax^2+bx+c=0# is given by #x=(-b+-sqrt(b^2-4ac))/(2a)#. Note that the quadratic equation is in its general form.

Hence converting the equation to its general form, the equation #2x^2=6x−2# this becomes #2x^2-6x+2=0# or dividing by #2#, #x^2-3x+1=0# and as such #a=1, b=-3# and #c=1#.

Hence, solution is given by #x=(-(-3)+-sqrt((-3)^2-4*1*1))/(2*1)# or

#x=(3+-sqrt(9-4))/2# or #x=(3+-sqrt(5))/2# i.e.

#x=(3+sqrt(5))/2# or #(3-sqrt(5))/2#