Find $f$ such that $f^{\prime }(x)=10x-3,f(2)=0$
$$f(x)=$$

Step 1 ：The problem is asking for a function $f(x)$ such that its derivative $f^{\prime }(x)$ is equal to $10x-3$ and $f(2)$ is equal to 0. This is a problem of finding the antiderivative (or integral) of a function, and then adjusting the constant of integration such that the function passes through the point (2, 0).

Step 2 ：The antiderivative of $10x-3$ is $5x^2-3x+C$, where $C$ is the constant of integration.

Step 3 ：We can find the value of $C$ by substituting $x=2$ and $f(2)=0$ into the equation and solving for $C$.

Step 4 ：Substituting these values into the equation gives us $C+14=0$, so $C=-14$.

Step 5 ：Substituting $C=-14$ back into the equation gives us the function $f(x)=5x^2-3x-14$.

Step 6 ：$\boxed{{\displaystyle f(x)=5x^2-3x-14}}$ is the function that satisfies the given conditions.