(9b+b)(4b-5)=0

Solution for (9b+b)(4b-5)=0 equation:

(9b+b)(4b-5)=0

We add all the numbers together, and all the variables

(+10b)(4b-5)=0

We multiply parentheses ..

(+40b^2-50b)=0

We get rid of parentheses

40b^2-50b=0

a = 40; b = -50; c = 0;
Δ = b2-4ac
Δ = -502-4·40·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-50}{2*40}=\frac{0}{80}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+50}{2*40}=\frac{100}{80}
=1+1/4
$