To solve the equation \(\sqrt{2x^2 + 3x + 7} = \sqrt{2x - 2}\), we can start by squaring both sides to eliminate the square roots. This gives us:

\[
(\sqrt{2x^2 + 3x + 7})^2 = (\sqrt{2x - 2})^2
\]

\[
2x^2 + 3x + 7 = 2x - 2
\]

Next, we need to move all terms to one side of the equation to set it to zero:

\[
2x^2 + 3x + 7 - 2x + 2 = 0
\]

\[
2x^2 + x + 9 = 0
\]

Now we have a quadratic equation. To solve it, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = 9\).

First, we calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot 9 = 1 - 72 = -71
\]

Since the discriminant is negative (\(-71\)), the quadratic equation \(2x^2 + x + 9 = 0\) has no real solutions. This means there are no real values of \(x\) that satisfy the original equation \(\sqrt{2x^2 + 3x + 7} = \sqrt{2x - 2}\).

Therefore, the equation has no real solutions.