How do you solve #3x^2+4x-3=0# using the quadratic formula?

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Answer 1 · 2016-07-14T09:30:08

#x=(-2+-2sqrt(13))/3#

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

solves equations of the type:

#ax^2+bx+c=0#

So, in this case, since

a=3; b=4: c=-3,

you have

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

#=(-4+-sqrt(16+36))/6#

#=(-4+-sqrt(52))/6#

#=(-4+-4sqrt(13))/6#

#=(-2+-2sqrt(13))/3#