(d+5)(d+3)=0

Solution for (d+5)(d+3)=0 equation:

(d+5)(d+3)=0

We multiply parentheses ..

(+d^2+3d+5d+15)=0

We get rid of parentheses

d^2+3d+5d+15=0

We add all the numbers together, and all the variables

d^2+8d+15=0

a = 1; b = 8; c = +15;
Δ = b2-4ac
Δ = 82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*1}=\frac{-10}{2}
=-5
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*1}=\frac{-6}{2}
=-3
$