請問換元法怎計這數

x^2 + 5x + 2[(3x^2 + 15x + 1)]^1/2 = 2
x^2 + 5x - 2 + 2[(3x^2 + 15x + 1)]^1/2 = 0
3x^2 + 15x - 6 + 6[(3x^2 + 15x + 1)]^1/2 = 0
3x^2 + 15x + 1 - 7 + 6[(3x^2 + 15x + 1)]^1/2 = 0
(3x^2 + 15x + 1) + 6[(3x^2 + 15x + 1)]^1/2 - 7 = 0
let y = [(3x^2 + 15x + 1)]^1/2
y^2 + 6y - 7 = 0
(y + 7)(y - 1 ) = 0
y = 1 or y = -7
[(3x^2 + 15x + 1)]^1/2 = 1 or [(3x^2 + 15x + 1)]^1/2 = -7(rejected)
3x^2 + 15x + 1 = 1
3x^2 + 15x = 0
3x(x + 5) = 0
x = 0 or x = -5

x = 0