(d+3)(d+11)=0

Solution for (d+3)(d+11)=0 equation:

(d+3)(d+11)=0

We multiply parentheses ..

(+d^2+11d+3d+33)=0

We get rid of parentheses

d^2+11d+3d+33=0

We add all the numbers together, and all the variables

d^2+14d+33=0

a = 1; b = 14; c = +33;
Δ = b2-4ac
Δ = 142-4·1·33
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-8}{2*1}=\frac{-22}{2}
=-11
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+8}{2*1}=\frac{-6}{2}
=-3
$