Find the point $(x, y)$, at which the graph of $y=5 x^{2}+3 x-3$ has a horizontal tangent. The function $y=5 x^{2}+3 x-3$ has a horizontal tangent at (Type an ordered pair. Type simplified fractions.)

Step 1 ：The function given is \(y=5 x^{2}+3 x-3\).

Step 2 ：The slope of the tangent line to the graph of a function at a particular point is given by the derivative of the function at that point.

Step 3 ：A horizontal tangent line has a slope of 0.

Step 4 ：Therefore, to find the point at which the graph of the function \(y=5 x^{2}+3 x-3\) has a horizontal tangent, we need to find the value of \(x\) for which the derivative of the function is 0.

Step 5 ：The derivative of the function \(y=5 x^{2}+3 x-3\) is \(y'=10x+3\).

Step 6 ：Setting \(y'=0\), we solve for \(x\) and get \(x=-3/10\).

Step 7 ：Substituting \(x=-3/10\) into the original function \(y=5 x^{2}+3 x-3\), we get \(y=-69/20\).

Step 8 ：Final Answer: The point at which the graph of \(y=5 x^{2}+3 x-3\) has a horizontal tangent is \(\boxed{\left(-\frac{3}{10}, -\frac{69}{20}\right)}\).