299. Find the sides of a triangle with an area of 84, if it is known that they are expressed by three consecutive integers.
299. Let the sides be $x-1, x, x+1$. The perimeter is $3x$, therefore,

$$
\begin{aligned}
& 84= \sqrt{\frac{3}{2} x \cdot \frac{1}{2} x \cdot\left(\frac{x}{2}+1\right)\left(\frac{x}{2}-1\right)}= \\
&=\sqrt{\frac{3 x^{2}(x+2)(x-2)}{16}} \\
& 84={ }_{4}^{x} \sqrt{3\left(x^{2}-4\right)} ; 336^{2}=x^{2} 3\left(x^{2}-4\right)
\end{aligned}
$$

$$
336 \cdot 112=\lambda^{4}-4 x^{2}
$$

or

$$
x^{4}-4 x^{2}-336 \cdot 112=0
$$

If $x^{2}=z$, then

$$
\begin{gathered}
z^{2}-4 z-336 \cdot 112=0 \\
z=2 \pm 2 \sqrt{1+33 b \cdot 28}=2 \pm 194
\end{gathered}
$$

The valid root is

$z_{1}=2+194=196 ; \quad 1=14$. Therefore, the sides of the triangle are $13,14,15$.