-10z^2+40z+50=0

Solution for -10z^2+40z+50=0 equation:

-10z^2+40z+50=0

a = -10; b = 40; c = +50;
Δ = b2-4ac
Δ = 402-4·(-10)·50
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{-(40)-60}{2*-10}=\frac{-100}{-20}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+60}{2*-10}=\frac{20}{-20}
=-1
$