3g+g(-6+3g)=1-g

Solution for 3g+g(-6+3g)=1-g equation:

3g+g(-6+3g)=1-g

We move all terms to the left:

3g+g(-6+3g)-(1-g)=0

We add all the numbers together, and all the variables

3g+g(3g-6)-(-1g+1)=0

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3g^2-2g-1=0

a = 3; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·3·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*3}=\frac{-2}{6}
=-1/3
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*3}=\frac{6}{6}
=1
$