(z-5)(z+4)=22

Solution for (z-5)(z+4)=22 equation:

(z-5)(z+4)=22

We move all terms to the left:

(z-5)(z+4)-(22)=0

We multiply parentheses ..

(+z^2+4z-5z-20)-22=0

We get rid of parentheses

z^2+4z-5z-20-22=0

We add all the numbers together, and all the variables

z^2-1z-42=0

a = 1; b = -1; c = -42;
Δ = b2-4ac
Δ = -12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-13}{2*1}=\frac{-12}{2}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+13}{2*1}=\frac{14}{2}
=7
$