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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

r^2+5r+6=0

a = 1; b = 5; c = +6;
Δ = b2-4ac
Δ = 52-4·1·6
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{-(5)-1}{2*1}=\frac{-6}{2}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*1}=\frac{-4}{2}
=-2
$