How do you find all the real and complex roots of #18x²+3x-1=0#?

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Answer 1 · 2016-02-01T11:56:43

here is how,

as #x# has the highest power of #2#, there are two values of #x#

we know,

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

#=(-3+-sqrt(3^2-4*18(-1)))/(2*18)#

#=(-3+-sqrt(9+72))/36#

#=(-3+-sqrt81)/36#

#=(-3+-9)/36#

using the + value,

#x=(-3+9)/36#

#=6/36#

#=1/6#

using the - value,
#x=(-3-9)/36#

#=-12/36#

#=-1/3#