1/7 (3x-5)^2=28 solve the following equation by factorisation
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Hence:-
- The value of X=19/3
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(3x-5)^2=28*7
9x^2-30x+25-196=0
3(3x^2-10x-57)=0
3x^2-19x+9x-57=0
x(3x-19)+3(3x-19)=0
(x+3)(3x-19)=0
x= -3,19/3
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