(z^2+5)-2(7+z)=-6

Solution for (z^2+5)-2(7+z)=-6 equation:

(z^2+5)-2(7+z)=-6

We move all terms to the left:

(z^2+5)-2(7+z)-(-6)=0

We add all the numbers together, and all the variables

(z^2+5)-2(z+7)-(-6)=0

We add all the numbers together, and all the variables

(z^2+5)-2(z+7)+6=0

We multiply parentheses

(z^2+5)-2z-14+6=0

We get rid of parentheses

z^2-2z+5-14+6=0

We add all the numbers together, and all the variables

z^2-2z-3=0

a = 1; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*1}=\frac{-2}{2}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2}
=3
$