y^4+16y^2-15=0

(y^2 + 8)^2 = 79

y^2 = -8 ± √79

y = ± √(-8 ± √79)

y⁴ +16 y² -15 = 0

( y² )² +16 y² -15 = 0

Substitute y² = x

x² +16 x -15 = 0

x² +16 x = 15

x² +16 x + 64 = 15 + 64

x² +16 x + 64 = 79

( x + 8 )² = 79

x + 8 = ± √79

x = ± √79 - 8

x₁= - √79 - 8

x₂= √79 - 8

Now:

y² = x

y² = x₁

y² = - √79 - 8

y = ± √( - √79 - 8 )

y = ± √ [ ( - 1 ) ∙ √( √79 + 8 ) ]

y = ± √( - 1 ) ∙ √( √79 + 8 )

y = ± i ∙ √( √79 + 8 )

y₁ = - i ∙ √( √79 + 8 )

y₂ = i ∙ √( √79 + 8 )

y² = x

y² = x₂

y² = √79 - 8

y = ± √(√79 - 8 )

y₃ = - √(√79 - 8 )

y₄ = √( √79 - 8 )

The solutions are:

- i ∙ √( √79 + 8 ) , i ∙ √( √79 + 8 ) , - √( √79 - 8 ) , √( √79 - 8 )

To solve the equation y^4 + 16y^2 - 15 = 0, we can use a substitution. Let z = y^2.

Substituting z into the equation, we get:

z^2 + 16z - 15 = 0.

Now we have a quadratic equation in terms of z. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.

Factoring:
The equation z^2 + 16z - 15 = 0 can be factored as (z + 5)(z - 3) = 0.

Setting each factor equal to zero gives us z + 5 = 0 or z - 3 = 0.

Solving for z, we find z = -5 or z = 3.

Now we substitute back y^2 for z to get the solutions for y:

For z = -5:
y^2 = -5
Since the square of a real number cannot be negative, there are no solutions for this case.

For z = 3:
y^2 = 3
Taking the square root of both sides, we obtain:
y = ±sqrt(3).

Therefore, the solutions for the equation y^4 + 16y^2 - 15 = 0 are y = sqrt(3) and y = -sqrt(3).