Let $f(x)=x^2+3x-10$. If we take $x_0=0$ as the starting point for the Newton method, then the value of $x_2$ is what?
I have tried but it appears I am wrong somewhere as the exercise from where I have taken this problem does not matches my answer of -190/33.
You have an error in your calculation for $x_1$. You should have $$ x_1 = 0 - \frac{f(0)}{f'(0)} = 0 - \frac{(0)^2 + 3(0) - 10}{2(0) + 3} = -\left(- \frac{10}{3} \right)=\color{red}{+} \frac{10}{3} $$ With this correction you should get $x_2 = \frac{190}{87} \approx 2.1839$, where you are indeed approximating the root $x=2$ of $x^2 + 3x -10$.