3g(g-7)=2(10+g)

Solution for 3g(g-7)=2(10+g) equation:

3g(g-7)=2(10+g)

We move all terms to the left:

3g(g-7)-(2(10+g))=0

We add all the numbers together, and all the variables

3g(g-7)-(2(g+10))=0

We multiply parentheses

3g^2-21g-(2(g+10))=0


We calculate terms in parentheses: -(2(g+10)), so:

2(g+10)

We multiply parentheses

2g+20

Back to the equation:

-(2g+20)


We get rid of parentheses

3g^2-21g-2g-20=0

We add all the numbers together, and all the variables

3g^2-23g-20=0

a = 3; b = -23; c = -20;
Δ = b2-4ac
Δ = -232-4·3·(-20)
Δ = 769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{769}}{2*3}=\frac{23-\sqrt{769}}{6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{769}}{2*3}=\frac{23+\sqrt{769}}{6}
$