2^3-2z^2+16z=0

Solution for 2^3-2z^2+16z=0 equation:

2^3-2z^2+16z=0

We add all the numbers together, and all the variables

-2z^2+16z+8=0

a = -2; b = 16; c = +8;
Δ = b2-4ac
Δ = 162-4·(-2)·8
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{5}}{2*-2}=\frac{-16-8\sqrt{5}}{-4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{5}}{2*-2}=\frac{-16+8\sqrt{5}}{-4}
$