Answer :
Let u = x², then we would have u² + 5u - 6 = 0
From here, we can factor it and get us (u-1)(u+6) = 0
So our solution for u is u = -6 or 1.
Now substitute u back to x².
x² = -6 or 1
x = ±√(-6) or ±√1
Since ±√(-6) is not real number, we ignore it.
Which leave us x = ±√1 = ±1
So our real solution is x = -1 or 1.
From here, we can factor it and get us (u-1)(u+6) = 0
So our solution for u is u = -6 or 1.
Now substitute u back to x².
x² = -6 or 1
x = ±√(-6) or ±√1
Since ±√(-6) is not real number, we ignore it.
Which leave us x = ±√1 = ±1
So our real solution is x = -1 or 1.
