0.4x^2+1.2x-3.6=0

Solution for 0.4x^2+1.2x-3.6=0 equation:

0.4x^2+1.2x-3.6=0

a = 0.4; b = 1.2; c = -3.6;
Δ = b2-4ac
Δ = 1.22-4·0.4·(-3.6)
Δ = 7.2
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{7.2}}{2*0.4}=\frac{-1.2-\sqrt{7.2}}{0.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{7.2}}{2*0.4}=\frac{-1.2+\sqrt{7.2}}{0.8}
$