-2(z+3)z=-z-4(z+2)

Solution for -2(z+3)z=-z-4(z+2) equation:

-2(z+3)z=-z-4(z+2)

We move all terms to the left:

-2(z+3)z-(-z-4(z+2))=0

We multiply parentheses

-2z^2-6z-(-z-4(z+2))=0


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

-z-4(z+2)

We add all the numbers together, and all the variables

-1z-4(z+2)

We multiply parentheses

-1z-4z-8

We add all the numbers together, and all the variables

-5z-8

Back to the equation:

-(-5z-8)


We get rid of parentheses

-2z^2-6z+5z+8=0

We add all the numbers together, and all the variables

-2z^2-1z+8=0

a = -2; b = -1; c = +8;
Δ = b2-4ac
Δ = -12-4·(-2)·8
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{65}}{2*-2}=\frac{1-\sqrt{65}}{-4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*-2}=\frac{1+\sqrt{65}}{-4}
$