(r+3)(r+1)=15

Solution for (r+3)(r+1)=15 equation:

(r+3)(r+1)=15

We move all terms to the left:

(r+3)(r+1)-(15)=0

We multiply parentheses ..

(+r^2+r+3r+3)-15=0

We get rid of parentheses

r^2+r+3r+3-15=0

We add all the numbers together, and all the variables

r^2+4r-12=0

a = 1; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*1}=\frac{-12}{2}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*1}=\frac{4}{2}
=2
$