How do you solve #2x^2 + x - 1 = 0 # using the quadratic formula?

1 Answer
Oct 15, 2015

Answer:

#x = 1/2, x = -1#

Explanation:

Standard Quadratic Equation :

#ax^2 + bx + c = 0#

Where,
a is the constant with x to the power 2.
b is the constant with x to the power 1.
c is the constant.

Standard Quadratic Formula:

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

Given Equation:
#2x^2 + x - 1 = 0#

In this Case:
a = 2
b = 1
c = -1

put these values in the quadratic formula.

#x = (-1 +- sqrt((1)^2 - 4(2)(-1))) / (2(2))#
#x = (-1 +- sqrt(1 + 8)) / 4#
#x = (-1 +- sqrt(9)) / 4#
#x = (-1 +-3 ) / 4#
#x = (-1 + 3) / 4 , (-1 -3) / 4#
#x = 2/4, x = -4/4#
#x = 1/2, x = -1#