How do you solve #3x^2+17x+20=0#?

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Answer 1 · 2018-05-05T02:45:41

#x = (-5)/3# or #x=-4#

#3x^2 +17x+20=0#

Can be factored by finding the factors for #a*c#

#3*20=60#

The factors of 60 are
#1*60#
#2*30#
#3*20#
#4*15#
#5*12#
#6*10#

The factor pattern that adds up to the middle term is #5+12 = 17#

Set the Polynomial up as

#3x^2 +12x +5x +20#

Now factor by grouping

#[3x^2 +12x][ +5x +20]#

#3x(x+4)+5(x+4)#

So the common factor is #(x+4)#

#(3x+5)(x+4)=0#

Set both factors equal to zero

#3x+5 = 0#

#x=-5/3#

#x+4=0#

#x=-4#