How do you use the quadratic formula to solve #2x^2-3x+2=0#?

1 Answer
Mar 30, 2018

Answer:

No real solutions. Imaginary solutions: #x=(3+-sqrt(7)i)/4#

Explanation:

The quadratic formula allows us to solve a quadratic equation in the form #ax^2+bx+c# for #x# using the following:

#x=(-b+-sqrt(b^4-4ac))/(2a)#. So, generally, there will be two solutions.

For #2x^2-3x+2, a=2, b=-3, c=2#, and so

#x=(3+-sqrt((-3)^2-(4)(2)(2)))/(2*2)=(3+-sqrt(9-16))/4=(3+-sqrt(-7))/4#

Due to the negative root, there will be no real solutions, but the imaginary solutions would be

#x=(3+-sqrt(7)i)/4#