4x2+20x+25
(2x+5)^2
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Assuming the missing signs are pluses, that factors to (2x + 5)(2x + 5)
362+60x+25 (6x+5)(6x+5)
(x-5i)(x+5i)
(6x + 5)(6x + 5) or (6x + 5)2
x2 + 25 doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: Zero plus or minus 5i times the square root of 1.x = 5ix = -5iwhere i is the square root of negative one.