-0.04x^2+0.8x+2=0

Solution for -0.04x^2+0.8x+2=0 equation:

-0.04x^2+0.8x+2=0

a = -0.04; b = 0.8; c = +2;
Δ = b2-4ac
Δ = 0.82-4·(-0.04)·2
Δ = 0.96
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{-(0.8)-\sqrt{0.96}}{2*-0.04}=\frac{-0.8-\sqrt{0.96}}{-0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{0.96}}{2*-0.04}=\frac{-0.8+\sqrt{0.96}}{-0.08}
$