-2(z+3)-z=-z-z(z+2)

Solution for -2(z+3)-z=-z-z(z+2) equation:

-2(z+3)-z=-z-z(z+2)

We move all terms to the left:

-2(z+3)-z-(-z-z(z+2))=0

We add all the numbers together, and all the variables

-1z-2(z+3)-(-z-z(z+2))=0

We multiply parentheses

-1z-2z-(-z-z(z+2))-6=0


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

-z-z(z+2)

We add all the numbers together, and all the variables

-1z-z(z+2)

We multiply parentheses

-z^2-1z-2z

We add all the numbers together, and all the variables

-1z^2-3z

Back to the equation:

-(-1z^2-3z)


We add all the numbers together, and all the variables

-(-1z^2-3z)-3z-6=0

We get rid of parentheses

1z^2+3z-3z-6=0

We add all the numbers together, and all the variables

z^2-6=0

a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2}
=-\frac{2\sqrt{6}}{2}
=-\sqrt{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2}
=\frac{2\sqrt{6}}{2}
=\sqrt{6}
$