How do you solve for #x# in #2x^2+4x-1=0# using the quadratic formula?

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Answer 1 · 2017-10-09T17:45:03

Let's see.

Let the quadratic equation be #ax^2 + bx + c= 0#

Then the quatratic formula for solving #x# is #rarr#

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

In the equation, #2x^2+4x-1=0#, #a=2,b=4,c=-1#.

Now, substituting the values we get,
#x =( -4 +-sqrt(4^2-4xx2xx(-1)))/(2*2)#

#:.x = (-4 +-sqrt(16 +8))/(4)#

#:.x = (-4 +-sqrt(24))/(4)#

#:.x = (-4 +-2sqrt(6))/(4)#

So, we get two real solutions for #x#:

#:.x=(sqrt(6)/2-1), x=-(sqrt(6)/2+1)# (Answer)

Hope it Helps :)