Answer: \(x=\frac12\), \(x=1\), \(x=3\).

At the points where the curve cuts the \(x\)-axis, \(y=0\). Therefore

\((2x-9)(x^2+5)+42=0.\)

Expand the expression:

\((2x-9)(x^2+5)+42=2x^3-9x^2+10x-45+42.\)

So

\(2x^3-9x^2+10x-3=0.\)

Use the factor theorem. Testing \(x=1\),

\(2(1)^3-9(1)^2+10(1)-3=0,\)

so \((x-1)\) is a factor. Dividing the cubic by \((x-1)\) gives

\(2x^3-9x^2+10x-3=(x-1)(2x^2-7x+3).\)

Now factorise the quadratic:

\(2x^2-7x+3=(2x-1)(x-3).\)

Hence

\((x-1)(2x-1)(x-3)=0.\)

Therefore the \(x\)-coordinates are

\(x=\frac12,\quad x=1,\quad x=3.\)