274. Solve the equations:

a) $\left(2 x^{2}+3 x-1\right)^{2}-5\left(2 x^{2}+3 x+3\right)+24=0$;
b) $(x-1)(x+3)(x+4)(x+8)=-96$

c) $(x-1)(x-2)(x-4)(x-8)=4 x^{2}$.
$\triangle$ a) Let's introduce the substitution: $2 x^{2}+3 x-1=y$. Then

$$
y^{2}-5(y+4)+24=0, \quad y^{2}-5 y+4=0
$$

From here, $y_{1}=4, y_{2}=1$. Knowing $y$, we find $x$.

b) Multiply the first factor by the fourth and the second by the third in the left part of the equation:

$$
\left(x^{2}+7 x-8\right)\left(x^{2}+7 x+12\right)=-96
$$

Assuming $x^{2}+7 x-8=y$, we get:

$$
y(y+20)=-96, \quad y^{2}+20 y+96=0 ; \quad y_{1}=-8, \quad y_{2}=-12
$$

It remains to calculate $x$ in both cases.

c) Here, too, in the left part of the equation, multiply the first factor by the fourth and the second by the third:

$$
\left(x^{2}-9 x+8\right)\left(x^{2}-6 x+8\right)=4 x^{2}
$$

But what next? Next, divide both sides of the equation by $x^{2}$, using the fact that $x=0$ is not a root of the equation:

$$
\left(x-9+\frac{8}{x}\right)\left(x-6+\frac{8}{x}\right)=4
$$

Introduce the substitution: $x-9+\frac{8}{x}=y$. We will have:

$$
y(y+3)=4, \quad y^{2}+3 y-4=0 ; \quad y_{1}=1, \quad y_{2}=-4
$$

In both cases, we find $x$.

Answer: a) $1,-2,1 / 2,-5 / 2$; b) $0,-7,(-1 \pm \sqrt{33}) / 2$; c) $5 \pm \sqrt{17}$.