(2g+9-5)-(6g-4+2);g=-2

Solution for (2g+9-5)-(6g-4+2);g=-2 equation:

(2g+9-5)-(6g-4+2)g=-2

We move all terms to the left:

(2g+9-5)-(6g-4+2)g-(-2)=0

We add all the numbers together, and all the variables

(2g+4)-(6g-2)g-(-2)=0

We add all the numbers together, and all the variables

(2g+4)-(6g-2)g+2=0

We multiply parentheses

-6g^2+(2g+4)+2g+2=0

We get rid of parentheses

-6g^2+2g+2g+4+2=0

We add all the numbers together, and all the variables

-6g^2+4g+6=0

a = -6; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-6)·6
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*-6}=\frac{-4-4\sqrt{10}}{-12}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*-6}=\frac{-4+4\sqrt{10}}{-12}
$