-0.32x^2+9.6x-40=0

Solution for -0.32x^2+9.6x-40=0 equation:

-0.32x^2+9.6x-40=0

a = -0.32; b = 9.6; c = -40;
Δ = b2-4ac
Δ = 9.62-4·(-0.32)·(-40)
Δ = 40.96
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{-(9.6)-\sqrt{40.96}}{2*-0.32}=\frac{-9.6-\sqrt{40.96}}{-0.64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9.6)+\sqrt{40.96}}{2*-0.32}=\frac{-9.6+\sqrt{40.96}}{-0.64}
$