(8g+1)(g-3)=0

Solution for (8g+1)(g-3)=0 equation:

(8g+1)(g-3)=0

We multiply parentheses ..

(+8g^2-24g+g-3)=0

We get rid of parentheses

8g^2-24g+g-3=0

We add all the numbers together, and all the variables

8g^2-23g-3=0

a = 8; b = -23; c = -3;
Δ = b2-4ac
Δ = -232-4·8·(-3)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*8}=\frac{-2}{16}
=-1/8
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*8}=\frac{48}{16}
=3
$