How do you solve #3x^2-10x+8=0#?

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Answer 1 · 2017-01-21T11:41:45

The answer is #S={2, 4/3}#

We compare this equation to

#ax^2+bx+c=0#

#3x^2-10x+8=0#

The discriminant is

#Delta=b^2-4ac=100-4*3*8=100-96=4#

As, #Delta>0#, there are 2 real roots

#x=(-b+-sqrt(Delta))/(2a)#

#x_1=(10+2)/6=2#

#x_2=(10-2)/6=8/6=4/3#

Answer 2 · 2017-01-21T12:28:28

#x=4/3# and# x=2#

#3x^2-10x+8=0#

#(3x-4)(x-2)#

#3x=4#
multiply both sides by#1/3#

#x=4/3#

#x-2=0#

#x=2#
substitute #x=4/3 and x=2#

#3(4/3)^2-10(4/3)+8=0#

#3 xx 16/9-40/3+8=0#

#cancel48^16/cancel9^3-40/3+8=0#

#5 1/3-13 1/3+8=0#

#-8+8=0#

#3(2)^2-10(2)+8=0#

#12-20+8=0#

#-8+8=0#