Answer: \(-1\leqslant x\leqslant3\).

First expand the left-hand side carefully:

\((3-x)(5x+8)=15x+24-5x^2-8x=-5x^2+7x+24.\)

So the inequality becomes

\(-5x^2+7x+24\geqslant9-3x.\)

Bring all terms to the left-hand side:

\(-5x^2+10x+15\geqslant0.\)

It is easier to factorise with a positive coefficient of \(x^2\). Multiplying by \(-1\) reverses the inequality:

\(5x^2-10x-15\leqslant0.\)

Divide by \(5\):

\(x^2-2x-3\leqslant0.\)

Now factorise:

\((x-3)(x+1)\leqslant0.\)

The critical values are \(x=-1\) and \(x=3\). Since the quadratic opens upwards, it is non-positive between the roots, including the roots themselves.

Therefore

\(-1\leqslant x\leqslant3.\)