(z+4)(z-7)=0

Solution for (z+4)(z-7)=0 equation:

(z+4)(z-7)=0

We multiply parentheses ..

(+z^2-7z+4z-28)=0

We get rid of parentheses

z^2-7z+4z-28=0

We add all the numbers together, and all the variables

z^2-3z-28=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$