.(g+4)-3g=1+g

Solution for .(g+4)-3g=1+g equation:

.(g+4)-3g=1+g

We move all terms to the left:

.(g+4)-3g-(1+g)=0

We add all the numbers together, and all the variables

.(g+4)-3g-(g+1)=0

We add all the numbers together, and all the variables

-3g+.(g+4)-(g+1)=0

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

g^2-4g=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
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{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$