Answer: (a) \(2\left(x+\frac54\right)^2-\frac18\); (b) \(\left(-\frac54,-\frac18\right)\); (c) \(-\frac94\lt x\lt-\frac14\).

For a quadratic expression, first find its roots or complete the square. The sign of the quadratic then follows from its shape and the position of the roots.

(a) Complete the square:

\(2x^2+5x+3=2\left(x^2+\frac52x\right)+3.\)

Half of \(\frac52\) is \(\frac54\), so

\(2x^2+5x+3 =2\left[\left(x+\frac54\right)^2-\frac{25}{16}\right]+3.\)

Hence

(b) The stationary point is therefore

\(\left(-\frac54,-\frac18\right).\)

(c) Solve

\(2\left(x+\frac54\right)^2-\frac18\lt\frac{15}{8}.\)

Then

\(2\left(x+\frac54\right)^2\lt2,\)

so

\(\left(x+\frac54\right)^2\lt1.\)

Thus

\(-1\lt x+\frac54\lt1.\)

Subtracting \(\frac54\),

\(-\frac94\lt x\lt-\frac14.\)