what are the answers for this 3x^4 + 10x^2 = 25?

you want x?

3x^4 + 10x^2 = 25
(3x^2 - 5)(x^2 + 5) = 0
x^2 = 5/3 or x^2 = -5

x = +-√(5/3) or x = +-i√5

3x^4 + 10x² - 25 = 0
Let y = x²

3y² + 10y - 25 = 0
y = [ - 10 ± √ (100 + 300) ] / 6
y = [ - 10 ± √ (400) ] / 6
y = [ - 10 ± 20 ] / 6
y = [ - 5 ± 10 ] / 3
y = 5/3 , y = - 5
x² = 5/3 , x² = 5 i²
x = ± √(5/3) , x = ± (√5) i

3x^4 + 10x^2 = 25
3x^4 + 10x^2 - 25 = 0
(3x^2 - 5)(x^2 + 5) = 0
3x^2 - 5 = 0
3x^2 = 5
x^2 = 5/3
x = sqrt 5/3
...OR...
x^2 + 5 = 0
x^2 = -5
x = sqrt -5

3x^4 + 10x^2 = 25
3x^4 + 10x^2 - 25 = 0
3x^4 + 15x^2 - 5x^2 - 25 = 0
(3x^4 + 15x^2) - (5x^2 + 25) = 0
3x^2(x^2 + 5) - 5(x^2 + 5) = 0
(x^2 + 5)(3x^2 - 5) = 0

x^2 + 5 = 0
x^2 = -5
x = ±√-5 (imaginary number)

3x^2 - 5 = 0
3x^2 = 5
x^2 = 5/3
x = ±√(5/3)

∴ x = ±√(5/3)

3x^4 + 10x^2 = 25

3x^4 + 10x^2 - 25 = 0
(3x^2 - 5)(x^2 + 5) = 0
so...
3x^2 - 5 = 0 or x^2 + 5 = 0
3x^2 = 5 or x^2 = -5
x = +/- sqrt(5/3) or x = +/- sqrt(5)i

sqrt(5/3),-sqrt(5/3),