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

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Answer 1 · 2015-07-15T12:49:10

#x=(-3+sqrt 33)/4,##(-3-sqrt 33)/4#

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

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

Substitute the given values into the formula.

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

#x=(-3+-sqrt(9+24))/4# =

#x=(-3+-sqrt33)/4#

Solve for #x#.

#x=(-3+sqrt33)/4#

#x=(-3-sqrt33)/4#