1-1.2x+0.2x^2=0

Solution for 1-1.2x+0.2x^2=0 equation:

1-1.2x+0.2x^2=0

a = 0.2; b = -1.2; c = +1;
Δ = b2-4ac
Δ = -1.22-4·0.2·1
Δ = 0.64
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{0.64}}{2*0.2}=\frac{1.2-\sqrt{0.64}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{0.64}}{2*0.2}=\frac{1.2+\sqrt{0.64}}{0.4}
$