4-1/3z=-7z+6

Solution for 4-1/3z=-7z+6 equation:

4-1/3z=-7z+6

We move all terms to the left:

4-1/3z-(-7z+6)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We get rid of parentheses

-1/3z+7z-6+4=0

We multiply all the terms by the denominator


7z*3z-6*3z+4*3z-1=0

Wy multiply elements

21z^2-18z+12z-1=0

We add all the numbers together, and all the variables

21z^2-6z-1=0

a = 21; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·21·(-1)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{30}}{2*21}=\frac{6-2\sqrt{30}}{42}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{30}}{2*21}=\frac{6+2\sqrt{30}}{42}
$