1/3g+7=12+1/6g

Solution for 1/3g+7=12+1/6g equation:

1/3g+7=12+1/6g

We move all terms to the left:

1/3g+7-(12+1/6g)=0


Domain of the equation: 3g!=0

g!=0/3

g!=0

g∈R



Domain of the equation: 6g)!=0

g!=0/1

g!=0

g∈R


We add all the numbers together, and all the variables

1/3g-(1/6g+12)+7=0

We get rid of parentheses

1/3g-1/6g-12+7=0

We calculate fractions


6g/18g^2+(-3g)/18g^2-12+7=0

We add all the numbers together, and all the variables

6g/18g^2+(-3g)/18g^2-5=0

We multiply all the terms by the denominator


6g+(-3g)-5*18g^2=0

Wy multiply elements

-90g^2+6g+(-3g)=0

We get rid of parentheses

-90g^2+6g-3g=0

We add all the numbers together, and all the variables

-90g^2+3g=0

a = -90; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-90)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-90}=\frac{-6}{-180}
=1/30
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-90}=\frac{0}{-180}
=0
$