(b-20)(2b-5)=0

Solution for (b-20)(2b-5)=0 equation:

(b-20)(2b-5)=0

We multiply parentheses ..

(+2b^2-5b-40b+100)=0

We get rid of parentheses

2b^2-5b-40b+100=0

We add all the numbers together, and all the variables

2b^2-45b+100=0

a = 2; b = -45; c = +100;
Δ = b2-4ac
Δ = -452-4·2·100
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-35}{2*2}=\frac{10}{4}
=2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+35}{2*2}=\frac{80}{4}
=20
$