How do you solve #2x^2+3x-4=0# using the quadratic formula?

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Answer 1 · 2016-12-07T01:49:11

#x=(-3+sqrt(41))/4,# #(-3-sqrt(41))/4#

#2x^2+3x-4=0# is a quadratic equation in the form #ax^2+bx+c#, where #a=2#, #b=3#, and #c=-4#.

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

Plug the known values into the equation and solve.

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

#x=(-3+-sqrt(9+32))/4#

#x=(-3+-sqrt(41))/4#

#x=(-3+sqrt(41))/4,# #(-3-sqrt(41))/4#