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Yes, this can be done this way:
(x-(-5))(x+a)=0, where a is any (positive or negative) number
This simplifies to: (x+5)(x+a) = 0
or written as quadratic: x2+(a+5)x+5a = 0
Solutions: x1 = -5 and x2 = -a,
so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )
and there will be only one solution: x1,2 = -5
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