To express \(2x^2 - 3x\) in the form \(a(x + b)^2 + c\), we complete the square:

Start with the equation:

\(y = 2x^2 - 3x\)

Factor out the 2 from the quadratic terms:

\(y = 2(x^2 - \frac{3}{2}x)\)

Complete the square inside the parentheses:

Take half of the coefficient of \(x\), square it, and add and subtract inside the bracket:

\(x^2 - \frac{3}{2}x = \left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\)

Substitute back into the equation:

\(y = 2\left(\left(x - \frac{3}{4}\right)^2 - \frac{9}{16}\right)\)

Distribute the 2:

\(y = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}\)

Thus, the equation in the form \(a(x + b)^2 + c\) is:

\(y = 2\left(x - \frac{3}{4}\right)^2 - \frac{9}{8}\)

The vertex form \(a(x - h)^2 + k\) gives the vertex \((h, k)\).

So, the vertex is \(\left(\frac{3}{4}, -\frac{9}{8}\right)\).