Determine the values of $a$ and $b$ which would result in the function $f(x)$ being differentiable at $x=5$.
\[
f(x)=\left\{\begin{array}{lll}
a x-8 & \text { for } \quad x< 5 \\
2 b x^{2}-5 x+7 & \text { for } \quad x \geq 5
\end{array}\right.
\]

Step 1 ：For the function to be differentiable at x=5, it must be continuous at x=5. So, we first find the values of a and b that make the function continuous at x=5.

Step 2 ：For continuity, the left-hand limit and right-hand limit must be equal at x=5. So, we have \(a(5)-8=2b(5)^2-5(5)+7\).

Step 3 ：Simplifying the equation, we get \(5a-8=50b-18\) or \(5a=50b+10\).

Step 4 ：Now, for differentiability, the left-hand derivative and right-hand derivative must be equal at x=5.

Step 5 ：The left-hand derivative is \(\frac{d}{dx}(ax-8)=a\) and the right-hand derivative is \(\frac{d}{dx}(2bx^2-5x+7)=4bx-5\).

Step 6 ：Setting the left-hand derivative equal to the right-hand derivative at x=5, we get \(a=4b(5)-5\) or \(a=20b-5\).

Step 7 ：Now we have a system of equations: \(\begin{cases} 5a=50b+10 \\ a=20b-5 \end{cases}\)

Step 8 ：Substituting the second equation into the first, we get \(5(20b-5)=50b+10\) or \(100b-25=50b+10\).

Step 9 ：Solving for b, we get \(50b=35\) or \(b=\frac{7}{10}\).

Step 10 ：Substituting the value of b back into the second equation, we get \(a=20(\frac{7}{10})-5\) or \(a=\frac{9}{2}\).

Step 11 ：Thus, the values of a and b are \(a=\frac{9}{2}\) and \(b=\frac{7}{10}\).

Step 12 ：\(\boxed{a=\frac{9}{2}, b=\frac{7}{10}}\)