How do you solve #2x^2 - 3x + 1 = 0#?

1 Answer
Apr 4, 2018

Answer:

#x = 1/2, 1#

Here's how I solved it:

Explanation:

You solve this by using the quadratic formula. This equation is written in standard form, or #ax^2 + bx + c = 0#. The quadratic formula is #x = (-b +- sqrt(b^2-4ac))/(2a)#.

So let's plug those numbers in and solve for #x#:
#x = (-(-3) +- sqrt((-3)^2 - 4(2)(1)))/(2(2))#

#x = (3 +- sqrt(9 - 8))/4#

#x = (3 +- sqrt(1))/4#

#x = (3 +- 1)/4#

So #x = (3+1)/4# and #(3-1)/4#.

#x = 4/4 -> x = 1#

#x = 2/4 -> x = 1/2#

#x = 1/2, 1#

Hope this helps!