-0.2x^2+3.2x-5=0

Solution for -0.2x^2+3.2x-5=0 equation:

-0.2x^2+3.2x-5=0

a = -0.2; b = 3.2; c = -5;
Δ = b2-4ac
Δ = 3.22-4·(-0.2)·(-5)
Δ = 6.24
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{-(3.2)-\sqrt{6.24}}{2*-0.2}=\frac{-3.2-\sqrt{6.24}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{6.24}}{2*-0.2}=\frac{-3.2+\sqrt{6.24}}{-0.4}
$