(G-4)*(g+4)=g2-16

Solution for (G-4)*(g+4)=g2-16 equation:

(-4)(G+4)=G2-16

We move all terms to the left:

(-4)(G+4)-(G2-16)=0

We add all the numbers together, and all the variables

-(+G^2-16)+(-4)(G+4)=0

We get rid of parentheses

-G^2+(-4)(G+4)+16=0

We multiply parentheses ..

-G^2+(-4G-16)+16=0

We add all the numbers together, and all the variables

-1G^2+(-4G-16)+16=0

We get rid of parentheses

-1G^2-4G-16+16=0

We add all the numbers together, and all the variables

-1G^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
$