50z^2+20z-16=0

Solution for 50z^2+20z-16=0 equation:

50z^2+20z-16=0

a = 50; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·50·(-16)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-60}{2*50}=\frac{-80}{100}
=-4/5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+60}{2*50}=\frac{40}{100}
=2/5
$