-0.5x^2+0.4x+8=0

Solution for -0.5x^2+0.4x+8=0 equation:

-0.5x^2+0.4x+8=0

a = -0.5; b = 0.4; c = +8;
Δ = b2-4ac
Δ = 0.42-4·(-0.5)·8
Δ = 16.16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.4)-\sqrt{16.16}}{2*-0.5}=\frac{-0.4-\sqrt{16.16}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{16.16}}{2*-0.5}=\frac{-0.4+\sqrt{16.16}}{-1}
$