- 4(y^2 - 3y + 2) \geq 0

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• yx^2 + y = 2x^2 - 3x + 1 \quad \Rightarrow \quad (y - 2)x^2 + 3x + (y - 1) = 0
• For \(x\) to be real, the discriminant of this quadratic must be non-negative:
• 3^2 - 4(y - 2)(y - 1) \geq 0
• - 4y^2 + 12y - 8 \geq 0
• -4y^2 + 12y + 1 \geq 0
• Multiply through by \(-1\) (reversing the inequality):

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