-z=12-(3/5)z

Solution for -z=12-(3/5)z equation:

-z=12-(3/5)z

We move all terms to the left:

-z-(12-(3/5)z)=0


Domain of the equation: 5)z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

-z-(12-(+3/5)z)=0

We add all the numbers together, and all the variables

-1z-(12-(+3/5)z)=0

We multiply all the terms by the denominator


-1z*5)z)-(12-(+3=0

Wy multiply elements

-5z^2+3=0

a = -5; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-5)·3
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*-5}=\frac{0-2\sqrt{15}}{-10}
=-\frac{2\sqrt{15}}{-10}
=-\frac{\sqrt{15}}{-5}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*-5}=\frac{0+2\sqrt{15}}{-10}
=\frac{2\sqrt{15}}{-10}
=\frac{\sqrt{15}}{-5}
$