Determine the number of real solutions of the system \( \left\{\begin{array}{l}y=2 x^{2}+1 \\ y=-x\end{array}\right. \) A. more than 2 B. 2 C. 1 D. 0

To find the number of real solutions to the system of equations: 1. \( y = 2x^2 + 1 \) 2. \( y = -x \) we can set the two equations equal to each other: \[ 2x^2 + 1 = -x \] Rearranging the equation gives us: \[ 2x^2 + x + 1 = 0 \] Now, we can determine the number of real solutions by calculating the discriminant (\(D\)) of the quadratic equation, given by \(D = b^2 - 4ac\), where \(a = 2\), \(b = 1\), and \(c = 1\): \[ D = 1^2 - 4 \cdot 2 \cdot 1 = 1 - 8 = -7 \] Since the discriminant is negative (\(D < 0\)), this means that the quadratic equation has no real solutions. Thus, the number of real solutions to the system of equations is: \[ \boxed{0} \] So the correct answer is D: 0.