how do i solve 4z^2=48z-80?

This is a quadratic, and Z will generally have 2 solutions (unless both are the same number in which case it will have 1)

What you need to do is factorize it first to make it more simple to deal with and we'll start by making it equal to zero by bringing everything to one side of the equation.

=> 4Z^2 -48z + 80 = 0
=> 4(Z^2 - 12z + 20) = 0

now we need to put the middle bit Into brackets. and we do this by trial and error really. what you need to construct is two brackets that when multiplied together give you Z^2 - 12z + 20.

so what ways can we multiply 2 numbers together to get 20?
1*20
2*10 <--- 10+2 = 12!
4*5

The fact that we can get 12 (as in 12z) from 2*10 is a giveaway that these are the numbers we want to use. However... we have a negative 12z so we'll want to use -10 and -2 (-10 + -2 = -12)


so we can write this as 4(z - 2)(z - 10) = 0
Try to expand these brackets back out again for practice, you'll find that they will make 4(Z^2 - 12z + 20) = 0 again.

now we can solve for Z because if A*B = 0 then either A or B is zero. (in this case we can now ignore the 4 because 4 does not equal 0)

so Either z -2 = 0
OR z - 10 = 0

so z = 10 OR 2
those are your answers, 10 and 2.

hope that helps a bit.

4z^2 = 48z - 80
z^2 = 12z - 20
z^2 - 12z + 20 = 0
z^2 - 2z - 10z + 20 = 0
(z^2 - 2z) - (10z - 20) = 0
z(z - 2) - 10(z - 2) = 0
(z - 2)(z - 10) = 0

z - 2 = 0
z = 2

z - 10 = 0
z = 10

∴ z = 2 , 10

4z^2-48z+80=0
z=[24 + & - (576-320)^1/2]/8=5 & 1

get everything on one side.
4z^2-48z+80=0
take a four out
4(z^2-12z+20)=0
factor
4(z-10)(z-2)
set the quantities equal to zero
z-10=0
z=10
and
z-2=0
z=2
so
z=10 and 2