A transformation matrix P transforms an object of area 10 cm2 to an image of area 60 cm2. If P = [[x, 4], [3, x+3]], find the possible values of x.

Area of Object A $= 10 \text{ cm}^2$
Area of Image B $= 60 \text{ cm}^2$
Transformation matrix $P = \begin{pmatrix} x & 4 \\ 3 & x+3 \end{pmatrix}$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

This leads to two cases.

Case 1: $x^2 + 3x - 12 = 6$

Factor the quadratic equation:

This gives two possible values for $x$:

Case 2: $x^2 + 3x - 12 = -6$

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=3$, $c=-6$.

This gives two possible values for $x$:

From Case 1: $x=3$ is positive. $x=-6$ is negative.
From Case 2: $\sqrt{33} \approx 5.74$.
$x = \frac{-3 + \sqrt{33}}{2} \approx \frac{-3 + 5.74}{2} = \frac{2.74}{2} = 1.37$ (positive).
$x = \frac{-3 - \sqrt{33}}{2} \approx \frac{-3 - 5.74}{2} = \frac{-8.74}{2} = -4.37$ (negative).

The positive values of $x$ are $\boxed{3 \text{ and } \frac{-3 + \sqrt{33}}{2}}$.