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

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

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

We multiply parentheses ..

(+4b^2+16b+5b+20)=0

We get rid of parentheses

4b^2+16b+5b+20=0

We add all the numbers together, and all the variables

4b^2+21b+20=0

a = 4; b = 21; c = +20;
Δ = b2-4ac
Δ = 212-4·4·20
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{-(21)-11}{2*4}=\frac{-32}{8}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*4}=\frac{-10}{8}
=-1+1/4
$