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

Solution for 2(b+4)(b-5)=0 equation:

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

We multiply parentheses ..

2(+b^2-5b+4b-20)=0

We multiply parentheses

2b^2-10b+8b-40=0

We add all the numbers together, and all the variables

2b^2-2b-40=0

a = 2; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·2·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*2}=\frac{-16}{4}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*2}=\frac{20}{4}
=5
$