0.2x^2+1.2x+1.6=0

Solution for 0.2x^2+1.2x+1.6=0 equation:

0.2x^2+1.2x+1.6=0

a = 0.2; b = 1.2; c = +1.6;
Δ = b2-4ac
Δ = 1.22-4·0.2·1.6
Δ = 0.16
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{-(1.2)-\sqrt{0.16}}{2*0.2}=\frac{-1.2-\sqrt{0.16}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{0.16}}{2*0.2}=\frac{-1.2+\sqrt{0.16}}{0.4}
$