2/3z+4=24-z

Solution for 2/3z+4=24-z equation:

2/3z+4=24-z

We move all terms to the left:

2/3z+4-(24-z)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

2/3z-(-1z+24)+4=0

We get rid of parentheses

2/3z+1z-24+4=0

We multiply all the terms by the denominator


1z*3z-24*3z+4*3z+2=0

Wy multiply elements

3z^2-72z+12z+2=0

We add all the numbers together, and all the variables

3z^2-60z+2=0

a = 3; b = -60; c = +2;
Δ = b2-4ac
Δ = -602-4·3·2
Δ = 3576
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:

$\sqrt{\Delta}=\sqrt{3576}=\sqrt{4*894}=\sqrt{4}*\sqrt{894}=2\sqrt{894}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-2\sqrt{894}}{2*3}=\frac{60-2\sqrt{894}}{6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+2\sqrt{894}}{2*3}=\frac{60+2\sqrt{894}}{6}
$