16x-0.5x^2=0

Solution for 16x-0.5x^2=0 equation:

16x-0.5x^2=0

a = -0.5; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-0.5)·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*-0.5}=\frac{-32}{-1}
=+32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-0.5}=\frac{0}{-1}
=0
$