1/3z-4=6z-38

Solution for 1/3z-4=6z-38 equation:

1/3z-4=6z-38

We move all terms to the left:

1/3z-4-(6z-38)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We get rid of parentheses

1/3z-6z+38-4=0

We multiply all the terms by the denominator


-6z*3z+38*3z-4*3z+1=0

Wy multiply elements

-18z^2+114z-12z+1=0

We add all the numbers together, and all the variables

-18z^2+102z+1=0

a = -18; b = 102; c = +1;
Δ = b2-4ac
Δ = 1022-4·(-18)·1
Δ = 10476
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:

$\sqrt{\Delta}=\sqrt{10476}=\sqrt{36*291}=\sqrt{36}*\sqrt{291}=6\sqrt{291}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(102)-6\sqrt{291}}{2*-18}=\frac{-102-6\sqrt{291}}{-36}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(102)+6\sqrt{291}}{2*-18}=\frac{-102+6\sqrt{291}}{-36}
$