Answer: \(x=3.17\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

At the intersections of the line and the curve,

\(4x^2-6x-5=1-4x\).

Rearrange:

\(4x^2-2x-6=0\).

Divide by 2:

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

Factorise:

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

So

\(x=-1\) or \(x=\frac32\).

Using \(y=1-4x\), the point with \(x=-1\) has \(y=5\), and the point with \(x=\frac32\) has \(y=-5\).

From the diagram, \(A=(-1,5)\) and \(B=\left(\frac32,-5\right)\).

The gradient of the curve is

\(\frac{dy}{dx}=8x-6\).

At \(B\), where \(x=\frac32\),

\(\frac{dy}{dx}=8\left(\frac32\right)-6=12-6=6\).

So the tangent at \(B\) has gradient 6.

Using point \(B\), its equation is

\(y+5=6\left(x-\frac32\right)\).

So

\(y=6x-14\).

The horizontal line through \(A\) is \(y=5\). At \(C\),

\(5=6x-14\).

Therefore

\(6x=19\),

so

\(x=\frac{19}{6}=3.166\ldots\).

Correct to 2 decimal places, the \(x\)-coordinate of \(C\) is \(3.17\).