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solve: (3x^2-7x-20)(2x-5)=0
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(3x^2-7x-20)(2x-5)=0 Factor: (3x^2-7x-20)
(x-4) (3x + 5) (2x - 5) = 0 x-4 = 0 and x = 4 3x +5 = 0 and x = -5/3 2x - 5 = 0 and x = 5/2 Check my work.
(3x^2-7x-20)(2x-5)=0 factorize it =(3x^2-12x+5x-20)(2x-5)=0 =(3x(x-4)+5(x-4))(2x-5)=0 =(x-4)(3x+5)(2x-5)=0 now when x-4=0,x=4 when 3x+5=0 then x=-5/3 and when 2x-5=0, x=5/2
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