To determine the number of real solutions for the given system of equations:
[tex]\[
\begin{array}{l}
y = 2x^2 + 3 \\
y = x + 2
\end{array}
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously.
1. Set the equations equal to each other. Since both expressions are equal to [tex]\(y\)[/tex], we can set them equal to each other to find [tex]\(x\)[/tex]:
[tex]\[
2x^2 + 3 = x + 2
\][/tex]
2. Rearrange the equation to form a standard quadratic equation:
[tex]\[
2x^2 + 3 - x - 2 = 0
\][/tex]
[tex]\[
2x^2 - x + 1 = 0
\][/tex]
3. Solve the quadratic equation [tex]\(2x^2 - x + 1 = 0\)[/tex]. We will use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = 1\)[/tex]:
4. Calculate the discriminant [tex]\(\Delta = b^2 - 4ac\)[/tex]:
[tex]\[
\Delta = (-1)^2 - 4(2)(1) = 1 - 8 = -7
\][/tex]
5. Interpret the discriminant:
Since the discriminant [tex]\(\Delta = -7\)[/tex] is less than zero, the quadratic equation [tex]\(2x^2 - x + 1 = 0\)[/tex] has no real solutions. This means that the parabola represented by [tex]\(y = 2x^2 + 3\)[/tex] and the line represented by [tex]\(y = x + 2\)[/tex] do not intersect at any real points.
Therefore, the correct answer is:
[tex]\[
\boxed{\text{D. zero}}
\][/tex]
