How do you solve #2x^2 + 2x – 1 = 0#?

1 Answer
Shantelle
Jul 18, 2018

#x = (-1 +- sqrt3)/2#

Explanation:

To solve this, we will use the quadratic formula. We know that this is in standard quadratic form, or #ax^2 + bx + c#, where #a = 2#, #b = 2#, and #c = -1#.

The quadratic formula is:
#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Now plug in the values of #a#, #b#, and #c# into the formula to find the value of #x#:
#x = (-2 +- sqrt(2^2 - 4(2)(-1)))/(2(2))#

#x = (-2 +- sqrt(4 + 8))/4#

#x = (-2 +- sqrt12)/4#

#x = (-2 +- 2sqrt3)/4#

#x = (-1 +- sqrt3)/2#

Hope this helps!

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