(z+1)(z+5)=16

Solution for (z+1)(z+5)=16 equation:

(z+1)(z+5)=16

We move all terms to the left:

(z+1)(z+5)-(16)=0

We multiply parentheses ..

(+z^2+5z+z+5)-16=0

We get rid of parentheses

z^2+5z+z+5-16=0

We add all the numbers together, and all the variables

z^2+6z-11=0

a = 1; b = 6; c = -11;
Δ = b2-4ac
Δ = 62-4·1·(-11)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{5}}{2*1}=\frac{-6-4\sqrt{5}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{5}}{2*1}=\frac{-6+4\sqrt{5}}{2}
$