it doesn't need to "come from" anywhere; as long as you do an operation to **both sides of the equals sign**, the equality is maintained.

In order to solve for a variable x, which is done by getting it alone on one side of the equals sign (i.e. 'isolate x'), you have to do the opposite of what is currently being done to it: for example when we have 5 being added to x in the equation x + 5 = 3, we do the opposite by subtracting 5 from both sides of the equation to get x = 3 - 5 = -2.

If x is being divided by 4 as in the equation x/4 = 2, we do the opposite of division and multiply both sides by 4 to get x = 2\*4 = 8. Remember that a fraction is just a division!

When multiple operations are being done to x, to make it easier for yourself *you go the reverse of the order of operations*: first "undo" the additions / subtractions, then "undo" the multiplications/divisions, then "undo" the powers. In your equation, we have:

1. Undo the subtraction of 5 by adding 5 to both sides: (x\^2)/4 - 5 = 4 -> (x\^2)/4 = 9

2. Undo the division by 4 by multiplying both sides by 4: (x\^2)/4 = 9 -> x\^2 = 36 (remember, **a fraction is just a division**, so you undo it via multiplication!)

3. Undo the power of 2 by taking the square root of both sides: x\^2 = 36 -> x = \[+ or -\] 6

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Edit: From your other comment it's clear that the context of this is quadratics; note that an equation is usually only solvable this was if you only have one power of x in play, i.e. only x\^2 in the equation; if both x\^2 and x (and/or other powers) are present x can't be isolated via this simple means, so more advanced methods like factoring, "completing the square", and the quadratic formula must be used. The quadratic formula is the "x=-b+-\[b2-4ac\] / 2a" you mention and you could certainly use it to solve this by taking a = 1/4, b = 0, c = -1, but since there is no x term (only x\^2) this would be an overcomplication.