2g(4g+3)+g(g-7)=

Solution for 2g(4g+3)+g(g-7)= equation:

2g(4g+3)+g(g-7)=

We move all terms to the left:

2g(4g+3)+g(g-7)-()=0

We add all the numbers together, and all the variables

2g(4g+3)+g(g-7)=0

We multiply parentheses

8g^2+g^2+6g-7g=0

We add all the numbers together, and all the variables

9g^2-1g=0

a = 9; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·9·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*9}=\frac{0}{18}
=0
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*9}=\frac{2}{18}
=1/9
$