17=3g(g+3)-g

Solution for 17=3g(g+3)-g equation:

17=3g(g+3)-g

We move all terms to the left:

17-(3g(g+3)-g)=0


We calculate terms in parentheses: -(3g(g+3)-g), so:

3g(g+3)-g

We add all the numbers together, and all the variables

-1g+3g(g+3)

We multiply parentheses

3g^2-1g+9g

We add all the numbers together, and all the variables

3g^2+8g

Back to the equation:

-(3g^2+8g)


We get rid of parentheses

-3g^2-8g+17=0

a = -3; b = -8; c = +17;
Δ = b2-4ac
Δ = -82-4·(-3)·17
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{67}}{2*-3}=\frac{8-2\sqrt{67}}{-6}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{67}}{2*-3}=\frac{8+2\sqrt{67}}{-6}
$