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

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

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

4b^2-1=0

a = 4; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·4·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*4}=\frac{-4}{8}
=-1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*4}=\frac{4}{8}
=1/2
$