15/3z=1z=

Solution for 15/3z=1z= equation:

15/3z=1z=

We move all terms to the left:

15/3z-(1z)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

-1z+15/3z=0

We multiply all the terms by the denominator


-1z*3z+15=0

Wy multiply elements

-3z^2+15=0

a = -3; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-3)·15
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-3}=\frac{0-6\sqrt{5}}{-6}
=-\frac{6\sqrt{5}}{-6}
=-\frac{\sqrt{5}}{-1}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-3}=\frac{0+6\sqrt{5}}{-6}
=\frac{6\sqrt{5}}{-6}
=\frac{\sqrt{5}}{-1}
$