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

Step 1 ：The problem is asking for a function \(f(x)\) such that its derivative \(f'(x)\) 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 ：\(\boxed{f(x) = 5x^2 - 3x - 14}\) is the function that satisfies the given conditions.