Solve the following system of equations.
\[
\left\{\begin{array}{l}
x^{2}+y^{2}=30 \\
x^{2}-y=10
\end{array}\right.
\]

If there is more than one solution, enter additional solutions If there is no real solution, use the "No solution" button.
\[
(x, y)=(\square, \square)
\]

Step 1 ：The system of equations is non-linear and consists of two equations. The first equation is a circle equation and the second one is a parabola equation.

Step 2 ：To solve this system, we can substitute the second equation into the first one to get a quadratic equation in terms of y.

Step 3 ：The new equation is \(y^{2} + y + 10 = 30\).

Step 4 ：Solving this quadratic equation gives us the values of y as -5 and 4.

Step 5 ：We can then substitute these values into the second equation to get the corresponding values of x.

Step 6 ：For y = -5, the solutions for x are \(-\sqrt{5}\) and \(\sqrt{5}\).

Step 7 ：For y = 4, the solutions for x are \(-\sqrt{14}\) and \(\sqrt{14}\).

Step 8 ：Thus, the solutions to the system of equations are \((-2.23606797749979, -5.00000000000000), (2.23606797749979, -5.00000000000000), (-3.74165738677394, 4.00000000000000), (3.74165738677394, 4.00000000000000)\).

Step 9 ：\(\boxed{(-2.23606797749979, -5.00000000000000), (2.23606797749979, -5.00000000000000), (-3.74165738677394, 4.00000000000000), (3.74165738677394, 4.00000000000000)}\)