Given the curve y = 4x2 − 6x + 5 and x=1.

Given the curve $y = 4x^2 - 6x + 5$ and $x=1$.
Substitute $x=1$ into the equation to find the corresponding $y$-value:

$$y = 4(1)^2 - 6(1) + 5$$ $$y = 4 - 6 + 5$$ $$y = 3$$

So, the point on the curve is $(1, 3)$.

Differentiate the curve's equation with respect to $x$ to find the gradient function $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{d}{dx}(4x^2 - 6x + 5)$$ $$\frac{dy}{dx} = 8x - 6$$

Now, evaluate the gradient of the tangent at $x=1$:

$$m_{\text{tangent}} = 8(1) - 6$$ $$m_{\text{tangent}} = 8 - 6$$ $$m_{\text{tangent}} = 2$$

The normal is perpendicular to the tangent. If $m_{\text{tangent}}$ is the gradient of the tangent, then the gradient of the normal $m_{\text{normal}}$ is given by $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$.

$$m_{\text{normal}} = -\frac{1}{2}$$

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (1, 3)$ and the gradient $m = -\frac{1}{2}$:

$$y - 3 = -\frac{1}{2}(x - 1)$$

Multiply both sides by 2 to eliminate the fraction:

$$2(y - 3) = -1(x - 1)$$ $$2y - 6 = -x + 1$$

Rearrange the equation into the standard form $Ax + By + C = 0$:

$$x + 2y - 6 - 1 = 0$$ $$x + 2y - 7 = 0$$

The equation of the normal to the curve at $x=1$ is $\boxed{x + 2y - 7 = 0}$.