Y=0.5x^2+0.2x+0.02

Solution for Y=0.5x^2+0.2x+0.02 equation:

=0.5Y^2+0.2Y+0.02

We move all terms to the left:

-(0.5Y^2+0.2Y+0.02)=0

We get rid of parentheses

-0.5Y^2-0.2Y-0.02=0

a = -0.5; b = -0.2; c = -0.02;
Δ = b2-4ac
Δ = -0.22-4·(-0.5)·(-0.02)
Δ = 6.9388939039072E-18
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.2)-\sqrt{6.9388939039072E-18}}{2*-0.5}=\frac{0.2-\sqrt{6.9388939039072E-18}}{-1}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.2)+\sqrt{6.9388939039072E-18}}{2*-0.5}=\frac{0.2+\sqrt{6.9388939039072E-18}}{-1}
$