How do you use the quadratic formula to solve #1/2x^2-3x+2=3x-1#?

1 Answer
Mar 28, 2017

Answer:

#x_1~~11.477#
#x_2~~0.523#

Explanation:

Making right side of the equation equate as 0, we end up with:
#1/2x^2-6x+3=0#

We can make it easier to visualise by multiplying both sides by 2,
#x^2-12x+6=0#

How many and what kinds of roots will we end up with?
#b^2-4ac=(-12)^2-4(1)(6)=120 > 0# (i.e. 2 real roots)

Let's find out what they are:

#x_(1,2)=-b/(2a)+-sqrt(b^2-4ac)/(2a)#

#x_1=-(-12)/(2*1)+sqrt(120)/(2*1)~~11.477#
#x_2=-(-12)/(2*1)-sqrt(120)/(2*1)~~0.523#