Question 31919
x^2-4x-5=0
Using (a): x^2-5x+x-5=0
=>x(x-5)+1(x-5)=0
=>(x-5)(x+1)=0
=>x=5 or x=-1;
Using (b): x^2-4x-5=0
=>(x^2-4x)=5
=>[x^2-4x+(4/2)^2-(4/2)^2]=5
=>(x-2)^2=5+4
=>(x-2)^2=9
=>(x-2)=3 or (x-2)=-3
=>x=5 or x=-1;
Using (c): x^2-4x-5=0
The roots are [(-b)+/-sqrt(b^2-4ac)]/2a
a=1, b=-4, and c=-5
Substitute the value of a,b,and c to get the roots using quadratic formula