Answer: \(x=1\) and \(x=-\frac52\).

The curve is

\(y=x^3+\dfrac32x^2-2x+1.\)

Differentiate:

\(\dfrac{\mathrm dy}{\mathrm dx}=3x^2+3x-2.\)

At \(x=0\), the gradient of the tangent is

\(-2.\)

Therefore the gradient of the normal is the negative reciprocal:

\(\dfrac12.\)

When \(x=0\), the point on the curve is

\((0,1).\)

So the equation of the normal is

\(y-1=\dfrac12x,\)

or

\(y=1+\dfrac12x.\)

To find where this normal cuts the curve again, solve

\(x^3+\dfrac32x^2-2x+1=1+\dfrac12x.\)

Subtract the right-hand side:

\(x^3+\dfrac32x^2-\dfrac52x=0.\)

Multiplying by \(2\),

\(2x^3+3x^2-5x=0.\)

Factorise:

\(x(2x^2+3x-5)=0.\)

Then

\(2x^2+3x-5=(2x+5)(x-1).\)

So the intersections have

\(x=0,\quad x=-\dfrac52,\quad x=1.\)

The value \(x=0\) is the original point of contact, so the two other \(x\)-coordinates are

\(x=1\quad\text{and}\quad x=-\dfrac52.\)