7/3z+10=2/9z-5

Solution for 7/3z+10=2/9z-5 equation:

7/3z+10=2/9z-5

We move all terms to the left:

7/3z+10-(2/9z-5)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: 9z-5)!=0

z∈R


We get rid of parentheses

7/3z-2/9z+5+10=0

We calculate fractions


63z/27z^2+(-6z)/27z^2+5+10=0

We add all the numbers together, and all the variables

63z/27z^2+(-6z)/27z^2+15=0

We multiply all the terms by the denominator


63z+(-6z)+15*27z^2=0

Wy multiply elements

405z^2+63z+(-6z)=0

We get rid of parentheses

405z^2+63z-6z=0

We add all the numbers together, and all the variables

405z^2+57z=0

a = 405; b = 57; c = 0;
Δ = b2-4ac
Δ = 572-4·405·0
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-57}{2*405}=\frac{-114}{810}
=-19/135
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+57}{2*405}=\frac{0}{810}
=0
$