(1/3)z+1/6=31/6

Solution for (1/3)z+1/6=31/6 equation:

(1/3)z+1/6=31/6

We move all terms to the left:

(1/3)z+1/6-(31/6)=0


Domain of the equation: 3)z!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+1/3)z+1/6-(+31/6)=0

We multiply parentheses

z^2+1/6-(+31/6)=0

We get rid of parentheses

z^2+1/6-31/6=0

We multiply all the terms by the denominator


z^2*6+1-31=0

We add all the numbers together, and all the variables

z^2*6-30=0

Wy multiply elements

6z^2-30=0

a = 6; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·6·(-30)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{5}}{2*6}=\frac{0-12\sqrt{5}}{12}
=-\frac{12\sqrt{5}}{12}
=-\sqrt{5}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{5}}{2*6}=\frac{0+12\sqrt{5}}{12}
=\frac{12\sqrt{5}}{12}
=\sqrt{5}
$