How do you find the real or imaginary solutions of the equation #6x^2+13x-5=0#?

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Answer 1 · 2016-11-28T10:56:34

The solutions are #S={1/3,-5/2}#

The simultaneous equations #ax^2+bx+c=0#

Our equation is #6x^2+13x-5=0#

We calculate the determinant,

#Delta=b^2-4ac#

#Delta=13^2-4*6*-5=289#

#Delta>0#, so we have 2 real roots

So,

#x=(-b+-sqrtDelta)/(2a)#

#=(-13+-sqrt289)/12#

#=(-13+-17)/12#

So, #x_1=-30/12=-5/2#

and #x_2=4/12=1/3#