(4b-12)(b-5)=0

Solution for (4b-12)(b-5)=0 equation:

(4b-12)(b-5)=0

We multiply parentheses ..

(+4b^2-20b-12b+60)=0

We get rid of parentheses

4b^2-20b-12b+60=0

We add all the numbers together, and all the variables

4b^2-32b+60=0

a = 4; b = -32; c = +60;
Δ = b2-4ac
Δ = -322-4·4·60
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8}{2*4}=\frac{24}{8}
=3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8}{2*4}=\frac{40}{8}
=5
$