(3g^2-g)+(3g^2-8g=4)=

Solution for (3g^2-g)+(3g^2-8g=4)= equation:

(3g^2-g)+(3g^2-8g=4)=

We move all terms to the left:

(3g^2-g)+(3g^2-8g-(4))=0

We get rid of parentheses

3g^2+3g^2-g-8g-4=0

We add all the numbers together, and all the variables

6g^2-9g-4=0

a = 6; b = -9; c = -4;
Δ = b2-4ac
Δ = -92-4·6·(-4)
Δ = 177
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{-(-9)-\sqrt{177}}{2*6}=\frac{9-\sqrt{177}}{12}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{177}}{2*6}=\frac{9+\sqrt{177}}{12}
$