∀x∈ℝ[≥0] ( sqrt(x)∈ℝ[≥0] ∧ sqrt(x)^2 = x ).  [property of square-root]
Given x,y∈ℝ:
	If x^2 = y:
		...
		y ≥ 0.
		...
		(x−sqrt(y))·(x+sqrt(y)) = 0.
		...
		[follow the same argument as in your own proof earlier]
Given x∈ℝ:
	If a·x^2+b·x+c = 0:
		...
		(2·a·x)^2+4·a·b·x+4·a·c = 0.
		...
		(2·a·x+b)^2 = b^2−4·a·c.
		...
		[use the above lemma]
		2·a·x+b = ... ∨ 2·a·x+b = ...
		...
		x = ... ∨ x = ...
	If x = ... ∨ x = ...:
		...
		a·x^2+b·x+c = 0.