50=z^2+14

Solution for 50=z^2+14 equation:

50=z^2+14

We move all terms to the left:

50-(z^2+14)=0

We get rid of parentheses

-z^2-14+50=0

We add all the numbers together, and all the variables

-1z^2+36=0

a = -1; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-1)·36
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-1}=\frac{-12}{-2}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-1}=\frac{12}{-2}
=-6
$