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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2r^2+4r+2r+4)=0

We get rid of parentheses

2r^2+4r+2r+4=0

We add all the numbers together, and all the variables

2r^2+6r+4=0

a = 2; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·2·4
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*2}=\frac{-8}{4}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*2}=\frac{-4}{4}
=-1
$