0=-0.02x^2+1x+0.5

Solution for 0=-0.02x^2+1x+0.5 equation:

0=-0.02x^2+1x+0.5

We move all terms to the left:

0-(-0.02x^2+1x+0.5)=0

We add all the numbers together, and all the variables

-(-0.02x^2+1x+0.5)=0

We get rid of parentheses

0.02x^2-1x-0.5=0

a = 0.02; b = -1; c = -0.5;
Δ = b2-4ac
Δ = -12-4·0.02·(-0.5)
Δ = 1.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{1.04}}{2*0.02}=\frac{1-\sqrt{1.04}}{0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1.04}}{2*0.02}=\frac{1+\sqrt{1.04}}{0.04}
$