How do you solve using the quadratic formula #3x^2 + 4x = 6#?

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Answer 1 · 2018-03-11T09:06:11

#x=(-4+-2sqrt22)/6#

The quadratic formula says that if we have a quadratic equation in the form:

#ax^2+bx+c=0#

The solutions will be:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

In our case, we have to subtract #6# from both sides to get it equal to #0#:

#3x^2+4x-6=0#

Now we can use the quadratic formula:
#x=(-4+-sqrt((-4)^2-4*3*-6))/(2*3)#

#x=(-4+-sqrt(16-(-72)))/6#

#x=(-4+-sqrt(88))/6=(-4+-sqrt(22*4))/6=(-4+-2sqrt22)/6#