If $\lim\limits_{x \to 1} \dfrac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$, then the quadratic equation having roots $a$ and $b$ is

Given limit is $\lim\limits_{x \to 1} \dfrac{\sin(3x^2 - 4x + 1) - x^2 + 1}{2x^3 - 7x^2 + ax + b} = -2$

Let $N(x) = \sin(3x^2 - 4x + 1) - x^2 + 1$ and $D(x) = 2x^3 - 7x^2 + ax + b$.

As $x \to 1$, $N(1) = \sin(0) - 1 + 1 = 0$.

For the limit to exist and be finite, the denominator must also approach $0$ as $x \to 1$.

$D(1) = 2(1)^3 - 7(1)^2 + a(1) + b = 0 \Rightarrow a + b = 5$

Applying L'Hopital's rule, we differentiate the numerator and the denominator:

$N'(x) = (6x - 4)\cos(3x^2 - 4x + 1) - 2x$

$D'(x) = 6x^2 - 14x + a$

As $x \to 1$, $N'(1) = 2\cos(0) - 2 = 0$.

Since $N'(1) = 0$ and the limit is a non-zero finite value, $D'(1)$ must also be $0$.

$D'(1) = 6(1)^2 - 14(1) + a = 0 \Rightarrow a = 8$

Substituting $a = 8$ into $a + b = 5$, we get $b = -3$.

Applying L'Hopital's rule again to verify the limit:

$N''(x) = -(6x - 4)^2\sin(3x^2 - 4x + 1) + 6\cos(3x^2 - 4x + 1) - 2$

$D''(x) = 12x - 14$

$\lim\limits_{x \to 1} \dfrac{N''(x)}{D''(x)} = \dfrac{0 + 6 - 2}{12 - 14} = \dfrac{4}{-2} = -2$, which matches the given limit.

The roots of the required quadratic equation are $a = 8$ and $b = -3$.

The sum of the roots is $a + b = 5$.

The product of the roots is $ab = 8 \times (-3) = -24$.

The quadratic equation is $x^2 - (a + b)x + ab = 0$, which gives $x^2 - 5x - 24 = 0$.

Answer: $x^2 - 5x - 24 = 0$