(2b+2)(4b+1)=0

Solution for (2b+2)(4b+1)=0 equation:

(2b+2)(4b+1)=0

We multiply parentheses ..

(+8b^2+2b+8b+2)=0

We get rid of parentheses

8b^2+2b+8b+2=0

We add all the numbers together, and all the variables

8b^2+10b+2=0

a = 8; b = 10; c = +2;
Δ = b2-4ac
Δ = 102-4·8·2
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*8}=\frac{-16}{16}
=-1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*8}=\frac{-4}{16}
=-1/4
$