What are the exact solutions of #x^2-x-4=0#?

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Answer 1 · 2016-11-16T11:18:10

The solutions are #S={2.56,-1.56}#

The equation is #x^2-x-4=0#

Let's calculate the discriminant

#Delta=b^2-4ac=(-1)^2-4*1*(-4)=17#

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

#x=(-b+-sqrtDelta)/(2a)=(1+-sqrt17)/2#

Therefore,

#x_1=(1+sqrt17)/2=2.56#

and #x_2=(1-sqrt17)/2=-1.56#