How do you solve #3x^2 + 4x = 2 #?

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Answer 1 · 2016-03-08T08:02:26

#x=(-2+sqrt10)/3# or #x=(-2-sqrt10)/3#

We may use quadratic formula to solve the equation #3x^2+4x=2# or #3x^2+4x-2=0#.

Quadratic formula gives the solution of equation #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#

As in #3x^2+4x-2=0#, #a=3#, #b=4# and #c=-2#,

#x=(-4+-sqrt(4^2-4xx3xx(-2)))/(2xx3)# or

#x=(-4+-sqrt(16+24))/6# or #(-4+-sqrt40)/6# or #(-4+-2sqrt10)/6# or

#x=(-2+-sqrt10)/3# i.e.

#x=(-2+sqrt10)/3# or #x=(-2-sqrt10)/3#