What is the range of a function and how do you solve for it?

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More generally, if we want to find the full range of y=x2+1, we can solve for x (taking the inverse of the function) to get x=√y−1. Then, the range of f(x) is simply the domain of √y−1, because these are all the values where there is some x-value with f(x)=y.

(Hope this helps!!!) Can I pls have a brainliest???

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Please solve: x^2+4x-8=0

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$x^2+4x-8=0\\\\\underbrace{x^2+2x\times2+2^2}_{use:(a+b)^2=a^2+2ab+b^2}-2^2-8=0\\\\(x+2)^2-4-8=0\\\\(x+2)^2-12=0\\\\(x+2)^2=12\iff x+2=\sqrt{12}\ or\ x+2=-\sqrt{12}\\\\x=\sqrt{4\times3}-2\ or\ x=-\sqrt{4\times3}-2\\\\x=\sqrt4\times\sqrt3-2\ or\ x=-\sqrt4\times\sqrt3-2\\\\\center\boxed{x=2\sqrt3-2\ or\ x=-2\sqrt3-2}$