(2/3)z=-24;z

Solution for (2/3)z=-24;z equation:

(2/3)z=-24z

We move all terms to the left:

(2/3)z-(-24z)=0


Domain of the equation: 3)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+2/3)z-(-24z)=0

We multiply parentheses

2z^2-(-24z)=0

We get rid of parentheses

2z^2+24z=0

a = 2; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·2·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*2}=\frac{-48}{4}
=-12
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*2}=\frac{0}{4}
=0
$