4-z^2+16=0

Solution for 4-z^2+16=0 equation:

4-z^2+16=0

We add all the numbers together, and all the variables

-1z^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}
=-\frac{2\sqrt{5}}{-1}
$
$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}
=\frac{2\sqrt{5}}{-1}
$