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Factor the 2nd degree polynomial under the radical √(4x^{2}-20x+25) = √(2x-5)(2x-5)=√(2x-5)^{2}

= ±(2x-5) and therefore there are two possible solutions; (2x-5) and -(2x-5)

Check the two possible solutions:

Knowing that (a-b)^{2} = a^{2} - 2ab + b^{2 }and (a+b)^{2 }= a^{2
}+ 2ab + b^{2}

Square the value ( 2x-5)^{2}= (2x)^{2} - (2)(2x)(5) + (5)^{2} = 4x^{2} -20x + 25

Square the value -(2x-5)^{2 }= (-2x+5)^{2 }= (-2x)^{2}+(2)(-2x)(5) + (5)^{2} = 4x^{2
}-20x + 25

Since both solutions lead to the same original polynomial under the radical, both solutions (2x-5)

and -(2x-5) are valid solutions.