3(2g+4)=6(g+2)g=

Solution for 3(2g+4)=6(g+2)g= equation:

3(2g+4)=6(g+2)g=

We move all terms to the left:

3(2g+4)-(6(g+2)g)=0

We multiply parentheses

6g-(6(g+2)g)+12=0


We calculate terms in parentheses: -(6(g+2)g), so:

6(g+2)g

We multiply parentheses

6g^2+12g

Back to the equation:

-(6g^2+12g)


We get rid of parentheses

-6g^2+6g-12g+12=0

We add all the numbers together, and all the variables

-6g^2-6g+12=0

a = -6; b = -6; c = +12;
Δ = b2-4ac
Δ = -62-4·(-6)·12
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-6}=\frac{-12}{-12}
=1
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-6}=\frac{24}{-12}
=-2
$