0.125x^2+16x=0

Solution for 0.125x^2+16x=0 equation:

0.125x^2+16x=0

a = 0.125; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·0.125·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*0.125}=\frac{-32}{0.25}
=-128
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*0.125}=\frac{0}{0.25}
=0
$