50z^2-32=0

Solution for 50z^2-32=0 equation:

50z^2-32=0

a = 50; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·50·(-32)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*50}=\frac{-80}{100}
=-4/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*50}=\frac{80}{100}
=4/5
$