How do you solve #4x^2+23x-10=10x + x^2# by using the quadratic formula?

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Answer 1 · 2017-03-07T12:29:38

#x= -5# or #x=2/3#

#4x^2+23x-10= 10x+x^2#

#3x^2+13x-10=0#

Because the question asks us use the quadratic formula:

#x= (-13+-sqrt(13^2-4xx3xx(-10)))/(2xx3)#

#= (-13+-sqrt(169 + 120))/(2xx3)#

#= (-13+-sqrt(289))/6#

#= (-13+-17)/6#

#= -5# or #2/3#

But also consider:

#3x^2+13x-10=0#

#(x+5)(3x-2) = 0 -> x= -5# or #2/3#

Therefore we could have arrived at the result without the quadratic formula in this case.