The polynomial $f(x)$ is such that $f(x)=6x^3+ax^2-50x+b$, where $a$ and $b$ are integers. It is given that $f(x)$ is divisible by $2x-3$ and that $f'(1)=6$.

1. Find the derivative of $f(x)$.

$f(x) = 6x^3 + ax^2 - 50x + b$
$f'(x) = 18x^2 + 2ax - 50$

2. Use the given condition $f'(1) = 6$ to find $a$.

$f'(1) = 18(1)^2 + 2a(1) - 50 = 6$
$18 + 2a - 50 = 6$
$2a - 32 = 6$
$2a = 38$
$a = 19$

3. Apply the factor theorem to set up an equation for $b$.

Since $f(x)$ is divisible by $2x - 3$, by the factor theorem, $f\left(\frac{3}{2}\right) = 0$.
$6\left(\frac{3}{2}\right)^3 + a\left(\frac{3}{2}\right)^2 - 50\left(\frac{3}{2}\right) + b = 0$

4. Substitute $a = 19$ and solve for $b$.

$6\left(\frac{27}{8}\right) + 19\left(\frac{9}{4}\right) - 75 + b = 0$
$\frac{81}{4} + \frac{171}{4} - 75 + b = 0$
$\frac{252}{4} - 75 + b = 0$
$63 - 75 + b = 0$
$-12 + b = 0$
$b = 12$

1. Substitute the values of $a$ and $b$ from part (a) and factorise $f(x)$ using the given linear factor

From part (a), $a = 19$ and $b = 12$.
$f(x) = 6x^3 + 19x^2 - 50x + 12$
Since $2x - 3$ is a factor, we can write:
$6x^3 + 19x^2 - 50x + 12 = (2x - 3)(3x^2 + cx - 4)$
Comparing the coefficients of $x^2$:
$-9 + 2c = 19$
$2c = 28 \implies c = 14$
So, $f(x) = (2x - 3)(3x^2 + 14x - 4)$

2. Solve the quadratic factor equal to zero

For $f(x) = 0$, either $2x - 3 = 0$ or $3x^2 + 14x - 4 = 0$.
$2x - 3 = 0 \implies x = 1.5$
For $3x^2 + 14x - 4 = 0$, using the quadratic formula:
$x = \frac{-14 \pm \sqrt{14^2 - 4(3)(-4)}}{2(3)}$
$x = \frac{-14 \pm \sqrt{196 + 48}}{6}$
$x = \frac{-14 \pm \sqrt{244}}{6}$

3. Evaluate the roots to 3 significant figures

$x = \frac{-14 + \sqrt{244}}{6} \approx 0.27008 \approx 0.270$
$x = \frac{-14 - \sqrt{244}}{6} \approx -4.9367 \approx -4.94$
The solutions are $x = 1.5$, $x = 0.270$, and $x = -4.94$.