Question

I donot know how to solve quadratic equations.
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

Answer 1

n^2-4n=32
n^2 - 4n - 32 = 0
n^2 - 8n + 4n - 32 = 0
n(n-8)+4(n-8)=0
(n-8)(n+4)=0
either n = 8 or n = - 4

Answer 2

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