-0.1x^2+0.4x+1.2=0

Solution for -0.1x^2+0.4x+1.2=0 equation:

-0.1x^2+0.4x+1.2=0

a = -0.1; b = 0.4; c = +1.2;
Δ = b2-4ac
Δ = 0.42-4·(-0.1)·1.2
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{-(0.4)-\sqrt{0.64}}{2*-0.1}=\frac{-0.4-\sqrt{0.64}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{0.64}}{2*-0.1}=\frac{-0.4+\sqrt{0.64}}{-0.2}
$