5/3z+3=7+z

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

5/3z+3=7+z

We move all terms to the left:

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


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

5/3z-z-7+3=0

We multiply all the terms by the denominator


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

Wy multiply elements

-3z^2-21z+9z+5=0

We add all the numbers together, and all the variables

-3z^2-12z+5=0

a = -3; b = -12; c = +5;
Δ = b2-4ac
Δ = -122-4·(-3)·5
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{51}}{2*-3}=\frac{12-2\sqrt{51}}{-6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{51}}{2*-3}=\frac{12+2\sqrt{51}}{-6}
$