-0.8x^2+2.4x+3.2=0

Solution for -0.8x^2+2.4x+3.2=0 equation:

-0.8x^2+2.4x+3.2=0

a = -0.8; b = 2.4; c = +3.2;
Δ = b2-4ac
Δ = 2.42-4·(-0.8)·3.2
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.4)-4}{2*-0.8}=\frac{-6.4}{-1.6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+4}{2*-0.8}=\frac{1.6}{-1.6}
=-1
$