-4+z2=16

Solution for -4+z2=16 equation:

-4+z2=16

We move all terms to the left:

-4+z2-(16)=0

We add all the numbers together, and all the variables

z^2-20=0

a = 1; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·1·(-20)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*1}=\frac{0-4\sqrt{5}}{2}
=-\frac{4\sqrt{5}}{2}
=-2\sqrt{5}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*1}=\frac{0+4\sqrt{5}}{2}
=\frac{4\sqrt{5}}{2}
=2\sqrt{5}
$