(z^2+7)-4(8+z)=-20

Solution for (z^2+7)-4(8+z)=-20 equation:

(z^2+7)-4(8+z)=-20

We move all terms to the left:

(z^2+7)-4(8+z)-(-20)=0

We add all the numbers together, and all the variables

(z^2+7)-4(z+8)-(-20)=0

We add all the numbers together, and all the variables

(z^2+7)-4(z+8)+20=0

We multiply parentheses

(z^2+7)-4z-32+20=0

We get rid of parentheses

z^2-4z+7-32+20=0

We add all the numbers together, and all the variables

z^2-4z-5=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$