(2r+1)(r+1)=0

Solution for (2r+1)(r+1)=0 equation:

(2r+1)(r+1)=0

We multiply parentheses ..

(+2r^2+2r+r+1)=0

We get rid of parentheses

2r^2+2r+r+1=0

We add all the numbers together, and all the variables

2r^2+3r+1=0

a = 2; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*2}=\frac{-4}{4}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*2}=\frac{-2}{4}
=-1/2
$