Factor.
\[
5 x^{2}-17 x+6
\]

Step 1 ：To factor the quadratic equation \(5x^{2}-17x+6\), we need to find two numbers that multiply to \(5*6=30\) (the product of the coefficient of \(x^{2}\) and the constant term) and add up to -17 (the coefficient of \(x\)).

Step 2 ：The numbers that satisfy these conditions are -15 and -2 because \(-15*-2=30\) and \(-15-2=-17\).

Step 3 ：So, we can write the middle term of the quadratic equation as the sum of the products of -15 and -2 with \(x\). This gives us: \(5x^{2}-15x-2x+6\).

Step 4 ：Next, we group the terms to factor by grouping: \(5x(x-3)-2(x-3)\).

Step 5 ：Now, we can see that \((x-3)\) is a common factor: \((x-3)(5x-2)\).

Step 6 ：So, the factored form of the quadratic equation \(5x^{2}-17x+6\) is \((x-3)(5x-2)\).

Step 7 ：To check our work, we can expand this product: \(x*5x - x*2 - 3*5x + 3*2 = 5x^{2} - 2x - 15x + 6 = 5x^{2} - 17x + 6\).

Step 8 ：This is the original expression, so our factorization is correct. The final answer is \(\boxed{(x-3)(5x-2)}\).