0.5x^2+0.2x-0.16=0

Solution for 0.5x^2+0.2x-0.16=0 equation:

0.5x^2+0.2x-0.16=0

a = 0.5; b = 0.2; c = -0.16;
Δ = b2-4ac
Δ = 0.22-4·0.5·(-0.16)
Δ = 0.36
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{-(0.2)-\sqrt{0.36}}{2*0.5}=\frac{-0.2-\sqrt{0.36}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{0.36}}{2*0.5}=\frac{-0.2+\sqrt{0.36}}{1}
$