What are the exact solutions of $x^{2} - x - 4 = 0$ $x^2-x-4=0$ ?

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Answer 1 · 2016-11-16T11:18:10

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

The equation is $x^{2} - x - 4 = 0$ $x^2-x-4=0$

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 $x_2=(1-sqrt17)/2=-1.56$

What are the exact solutions of x^2-x-4=0x2−x−4=0x^2-x-4=0?