(3g-11)(g-7)=0

Solution for (3g-11)(g-7)=0 equation:

(3g-11)(g-7)=0

We multiply parentheses ..

(+3g^2-21g-11g+77)=0

We get rid of parentheses

3g^2-21g-11g+77=0

We add all the numbers together, and all the variables

3g^2-32g+77=0

a = 3; b = -32; c = +77;
Δ = b2-4ac
Δ = -322-4·3·77
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-10}{2*3}=\frac{22}{6}
=3+2/3
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+10}{2*3}=\frac{42}{6}
=7
$