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Can anybody explain to solve the equation n^2-4n=32.In this equation we should find the value of n.Please expalin step by step along with basics of this topic

Here our eq. is n^2-4n=32 so n^2-4n-32=0 now we should split the multiple of 32 in such a way that its addition or subtraction comes out to be 4 now multiple of 32=2*16 =4*8 and many more but for 4*8=32 8-4=4 so n^2-[8n-4n]-32=0 so n^2-8n+4n-32=0 taking n common from first two terms and 4 from nest two terms we get n(n-8)+4(n-8)=0 taking n-8 common (n-8)(n+4)=0 as this is equal to zero any one of them should be zero so if (n-8) is zero than n-8=0 so n=8......................1 or if n+4 is zero than n+4=0 so n=-4..........................2 so the value of n are 8 or -4