2x^2-3x-2=0
ax^2 -bx-c=0  for example

x=-b(+-)Sqroot(4ac)                 
           2a
     
now x= -(-3)(+-)sqroot{4[2](-2)]}         
                         2x2


x= (+3)(+-)sqroot[16]
             4


x has two values one is in positive and one is in negative
because square root has positive and negative value

x=+7/4        ,     x= _1/4

It is hard for me to write it down here so I have found for you this link in wikipedia, that explains exactly what is the way to solve this equation

http://en.wikipedia.org/wiki/Quadratic_equation

hope I was helpful

2/5 x 3.4

2x^2-3x-2=0

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

Here

a=2
b =-3
c=-2

Putting the value in equation u will get

x = (3 + sqt((-3)^2 - 4(2)(2))) / 2(2)

x = (3 + sqt(9 + 16)) / 4

x = (3 + sqt(25)) / 4

x = (3 + 5) / 4

x = (3 + 5) / 4  or  x = (3 - 5)) / 4

x = 8/ 4  or  x = -2/ 4

x = 2  or  x = -1/2 or -0.5  >>answer