Let f (x) be an even function defined on R, and if x ≥ 0, f (x) = x2-2x-3, the root of the equation f (x) = 2a-3 (a ∈ R) is discussed Please answer as soon as possible!

Let x0, f (- x) = (- x) &# 178; - 2 (- x) - 3 = x & # 178; + 2x-3
When x0, f (x) = 2a-3, there are two intersections, x = 1 + √ (2a + 1), x = - 1 - √ (2a + 1)
in summary:
When A0, x = 1 + √ (2a + 1), x = - 1 - √ (2a + 1)

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