(1/3)(6g-9)=7-2g

Solution for (1/3)(6g-9)=7-2g equation:

(1/3)(6g-9)=7-2g

We move all terms to the left:

(1/3)(6g-9)-(7-2g)=0


Domain of the equation: 3)(6g-9)!=0

g∈R


We add all the numbers together, and all the variables

(+1/3)(6g-9)-(-2g+7)=0

We get rid of parentheses

(+1/3)(6g-9)+2g-7=0

We multiply parentheses ..

(+6g^2+1/3*-9)+2g-7=0

We multiply all the terms by the denominator


(+6g^2+1+2g*3*-9)-7*3*-9)=0

We add all the numbers together, and all the variables

(+6g^2+1+2g*3*-9)=0

We get rid of parentheses

6g^2+2g*3*+1-9=0

We add all the numbers together, and all the variables

6g^2+2g*3*-8=0

Wy multiply elements

6g^2+6g^2-8=0

We add all the numbers together, and all the variables

12g^2-8=0

a = 12; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·12·(-8)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*12}=\frac{0-8\sqrt{6}}{24}
=-\frac{8\sqrt{6}}{24}
=-\frac{\sqrt{6}}{3}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*12}=\frac{0+8\sqrt{6}}{24}
=\frac{8\sqrt{6}}{24}
=\frac{\sqrt{6}}{3}
$