How should i solve -x^4 + 6x^2 - 5 = 0?

Here is the process of solving for x
First multiply the equation by -1

x^4 - 6x^2 + 5 = 0

Now we need to factor the trinomial

(x^2 - 5)(x^2 - 1) = 0

For the product of the two parenthesis to be = 0
either one has to be = 0

Either
x^2 - 5 = 0

x^2 = 5

x = ±√5

Or

x^2 - 1 = 0

x = ± 1

-x^4 + 6x^2 - 5 = 0
-x^4 + 5x^2 + x^2 - 5 = 0
(-x^4 + 5x^2) + (x^2 - 5) = 0
-x^2(x^2 - 5) + 1(x^2 - 5) = 0
(x^2 - 5)(-x^2 + 1) = 0

x^2 - 5 = 0
x^2 = 5
x = ±√5

-x^2 + 1 = 0
-x^2 = -1
x^2 = -1/-1
x^2 = 1
x = ±√1
x = ±1

∴ x = ±√5 , ±1

√(-1) = i

x² = -5
x = ±√(-5)
= ±√(-1)√5
= ±i√5

Similarly, x = ±i

-x^4 + 6x^2 - 5 = 0-----If i don't mistake......(-x^4)
(-x^2 + 1) (x^2 - 5) = 0

-x^2 = -1
x^2 = 1
x = (1)^1/2
x = (+/-) 1


x^2 = 5
x = (5)^1/2
x = -/+ (5)^1/2


Ok.....

It will be well to start at the beginning. Change all the signs (i.e., multiply through by -1) to get x^4 -6x^2 +5 = 0. This factors to give (x^2 -1)(x^2 -5) = 0, so x^2 = +1 or +5. So x = +1, -1, +sqrt(5), or -sqrt(5).

-x^4 + 6x^2 - 5 = 0

Add 5 to both sides.
-x^4 + 6x^2 = 5

Group the LHS.
(-x^4 + 6x^2) = 5

Factor out -1 on the LHS.
-1(x^4 - 6x^2) = 5

Complete the square.
-1(x^4 - 6x^2 + _____) = 5 + -1(_____)

Take the coefficient of the x^2 term: -6
Divide it by 2: -6/2 = -3
Square it: (-3)^2 = 9
Add this to both sides.
-1(x^4 - 6x^2 + 9) = 5 + -1(9)
-1(x^4 - 6x^2 + 9) = 5 + -9
-1(x^4 - 6x^2 + 9) = -4

Divide by sides by -1.
-1(x^4 - 6x^2 + 9) / -1 = -4/-1
x^4 - 6x^2 + 9 = 4
(x^2 - 3)^2 = 4

Take the square root of both sides.
√(x^2 - 3)^2 = √4
x^2 - 3 = ± 2
x^2 = ± 2 + 3
x^2 = 2 + 3 or x^2 = 2 - 3
x^2 = 5 or x^2 = -1

Take the square root of both sides.
√x^2 = √5
x = ± √5

√x^2 = √(-1)
x = i

ANSWER: x = ± √5 or i

x^2 = -5
x = sqrt(-5) = sqrt(5) i or no real roots.