3z=(-7/3z)

Solution for 3z=(-7/3z) equation:

3z=(-7/3z)

We move all terms to the left:

3z-((-7/3z))=0


Domain of the equation: 3z))!=0

z!=0/1

z!=0

z∈R


We multiply all the terms by the denominator


3z*3z))-((-7=0

Wy multiply elements

9z^2-7=0

a = 9; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·9·(-7)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*9}=\frac{0-6\sqrt{7}}{18}
=-\frac{6\sqrt{7}}{18}
=-\frac{\sqrt{7}}{3}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*9}=\frac{0+6\sqrt{7}}{18}
=\frac{6\sqrt{7}}{18}
=\frac{\sqrt{7}}{3}
$