## Task B-2.1.

Determine all real solutions of the equation

$$
(x+3)(x+5)=4+\frac{2021}{(x+1)(x+7)}
$$
## Solution.

The given equation is equivalent to the equation

$$
x^{2}+8 x+15=4+\frac{2021}{x^{2}+8 x+7}, x \neq-1, x \neq-7
$$

By introducing the substitution $x^{2}+8 x+7=t$ we get the equation

$$
t+8=4+\frac{2021}{t}
$$

or

$$
t^{2}+4 t-2021=0
$$

whose solutions are $t_{1}=-47$ and $t_{2}=43$.

The equation $x^{2}+8 x+7=-47$, or $x^{2}+8 x+54=0$ has no real solutions, as its discriminant is less than zero.

The equation $x^{2}+8 x+7=43$, or $x^{2}+8 x-36=0$ has two distinct real solutions $x_{1,2}=-4 \pm 2 \sqrt{13}$.