(6b-7)(b-4)=0

Solution for (6b-7)(b-4)=0 equation:

(6b-7)(b-4)=0

We multiply parentheses ..

(+6b^2-24b-7b+28)=0

We get rid of parentheses

6b^2-24b-7b+28=0

We add all the numbers together, and all the variables

6b^2-31b+28=0

a = 6; b = -31; c = +28;
Δ = b2-4ac
Δ = -312-4·6·28
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-17}{2*6}=\frac{14}{12}
=1+1/6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+17}{2*6}=\frac{48}{12}
=4
$