(3b-4)(8b-7)=0

Solution for (3b-4)(8b-7)=0 equation:

(3b-4)(8b-7)=0

We multiply parentheses ..

(+24b^2-21b-32b+28)=0

We get rid of parentheses

24b^2-21b-32b+28=0

We add all the numbers together, and all the variables

24b^2-53b+28=0

a = 24; b = -53; c = +28;
Δ = b2-4ac
Δ = -532-4·24·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-53)-11}{2*24}=\frac{42}{48}
=7/8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+11}{2*24}=\frac{64}{48}
=1+1/3
$