How do you solve #x^2=x+6#?

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Answer 1 · 2016-02-04T12:14:44

#x^2=x+6#

#rarrx^2-x=6#

#rarrx^2-x-6=0#

Now this is a Quadratic equation (in form #ax^2+bx^2+c#)

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

In this case #a=1,b=-1,c=-6#

#rarrx=(-(-1)+-sqrt((-1)^2-4(1)(-6)))/(2(1))#

#rarrx=(1+-sqrt(1-(-24)))/2#

#rarrx=(1+-sqrt(1+24))/2#

#rarrx=(1+-sqrt25)/2#

#rarrx=(1+-5)/2#

Now we are in a stage in which we take the two answers:

#rarrx=(1+5)/2,(1-5)/2#

#rarrx=6/2,-4/2#

#rarrx=3,-2#

Answer 2 · 2016-02-04T12:19:58

This is another way of solving this equation:

#x^2=x+6#

#rarrx^2-x-6=0#

Factor:

#rarr(x-3)(x+2)=0#

#rarrx=3,-2#

=)