5z-5/7z=3

Solution for 5z-5/7z=3 equation:

5z-5/7z=3

We move all terms to the left:

5z-5/7z-(3)=0


Domain of the equation: 7z!=0

z!=0/7

z!=0

z∈R


We multiply all the terms by the denominator


5z*7z-3*7z-5=0

Wy multiply elements

35z^2-21z-5=0

a = 35; b = -21; c = -5;
Δ = b2-4ac
Δ = -212-4·35·(-5)
Δ = 1141
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{1141}}{2*35}=\frac{21-\sqrt{1141}}{70}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{1141}}{2*35}=\frac{21+\sqrt{1141}}{70}
$