z^2+2z=2(z+32)

Solution for z^2+2z=2(z+32) equation:

z^2+2z=2(z+32)

We move all terms to the left:

z^2+2z-(2(z+32))=0


We calculate terms in parentheses: -(2(z+32)), so:

2(z+32)

We multiply parentheses

2z+64

Back to the equation:

-(2z+64)


We get rid of parentheses

z^2+2z-2z-64=0

We add all the numbers together, and all the variables

z^2-64=0

a = 1; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·1·(-64)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*1}=\frac{-16}{2}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*1}=\frac{16}{2}
=8
$