0=2x^2-16x+0.5

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

0=2x^2-16x+0.5

We move all terms to the left:

0-(2x^2-16x+0.5)=0

We add all the numbers together, and all the variables

-(2x^2-16x+0.5)=0

We get rid of parentheses

-2x^2+16x-0.5=0

a = -2; b = 16; c = -0.5;
Δ = b2-4ac
Δ = 162-4·(-2)·(-0.5)
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{7}}{2*-2}=\frac{-16-6\sqrt{7}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{7}}{2*-2}=\frac{-16+6\sqrt{7}}{-4}
$