(z+1)(z-7)=0

Solution for (z+1)(z-7)=0 equation:

(z+1)(z-7)=0

We multiply parentheses ..

(+z^2-7z+z-7)=0

We get rid of parentheses

z^2-7z+z-7=0

We add all the numbers together, and all the variables

z^2-6z-7=0

a = 1; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·1·(-7)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-8}{2*1}=\frac{-2}{2}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+8}{2*1}=\frac{14}{2}
=7
$