8z=4z(2z+1)

Solution for 8z=4z(2z+1) equation:

8z=4z(2z+1)

We move all terms to the left:

8z-(4z(2z+1))=0


We calculate terms in parentheses: -(4z(2z+1)), so:

4z(2z+1)

We multiply parentheses

8z^2+4z

Back to the equation:

-(8z^2+4z)


We get rid of parentheses

-8z^2+8z-4z=0

We add all the numbers together, and all the variables

-8z^2+4z=0

a = -8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-8)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-8}=\frac{-8}{-16}
=1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-8}=\frac{0}{-16}
=0
$