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30 January, 02:10

What are the real roots of x2-7*+10=0

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  1. 30 January, 04:33
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    Is that x^2-7x+10? If so, use the AC method by multiplying the leading coefficient with the constant. Leading coefficient is 1 and constant is 10. 1x10 = 10 and then you have to multiply two numbers to get 10 and add two number that give you - 7. in this case it's - 5 and - 2. so that means x^2-5x-2x+10 and then factor out the first two x (x-5) and factor out the last two - 2 (x-5) combine the two (x-2) (x-5) = 0 solve for both x values x = 2 and 5.
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