2(2x^2+1)^2+7(2x^2+1)+3=0

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To solve this quadratic equation, let's make a substitution:
Let y = 2x^2 + 1.

Now our equation becomes:
2y^2 + 7y + 3 = 0

Next, we can factor this quadratic equation:
2y^2 + 6y + y + 3 = 0
2y(y + 3) + 1(y + 3) = 0
(y + 3)(2y + 1) = 0

Now, we can solve for y:
y + 3 = 0 or 2y + 1 = 0
y = -3 2y = -1
y = -1/2

Now substitute back to find x:
For y = -3:
2x^2 + 1 = -3
2x^2 = -4
x^2 = -2
x = ±√(-2)

For y = -1/2:
2x^2 + 1 = -1/2
2x^2 = -3/2
x^2 = -3/4
x = ±√(-3/4)

Therefore, the solutions to the equation 2(2x^2+1)^2+7(2x^2+1)+3=0 are:
x = ±√(-2) and x = ±√(-3/4).