(z^2+5)-3(7+z)=2

Solution for (z^2+5)-3(7+z)=2 equation:

(z^2+5)-3(7+z)=2

We move all terms to the left:

(z^2+5)-3(7+z)-(2)=0

We add all the numbers together, and all the variables

(z^2+5)-3(z+7)-2=0

We multiply parentheses

(z^2+5)-3z-21-2=0

We get rid of parentheses

z^2-3z+5-21-2=0

We add all the numbers together, and all the variables

z^2-3z-18=0

a = 1; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*1}=\frac{-6}{2}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*1}=\frac{12}{2}
=6
$