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

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Answer 1 · 2018-08-03T07:06:54

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

#3x^2 + 4x = 2#

First, make one side equal to zero by subtracting #color(blue)2# from both sides:
#3x^2 + 4x - 2 = 0#

The quadratic formula is #x = (-b +- sqrt(b^2 - 4ac))/(2a)#, where from our equation we know that #a = 3, b = 4, and c = -2#.

Plug them into the formula:
#x = (-4 +- sqrt(4^2 - 4(3)(-2)))/(2(3))#

#x = (-4 +- sqrt(16 + 24))/6#

#x = (-4 +- sqrt40)/6#

#x = (-4 +- 2sqrt10)/6#

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

Hope this helps!