Let's first write the equation given
x²+6x-13=0
They have told us to use "Completing the square" method.
So we need write x²+6x-13 as a whole square.
x²+6x-13=0
x²+(2*3*x)-9-4 = 0
x²+(2*3*x)-3²=4
Now we can write the LHS as:
(x - 3)²=2²
Hence,
x-3 can either be 2 or -2
Now,
x - 3=2
x=5
Or,
x - 3=-2
x=1
Therefore now we know that x can either be 1 or 5. So the correct answer is Option B.
