-0.1x^2+0.8x+6=0

Solution for -0.1x^2+0.8x+6=0 equation:

-0.1x^2+0.8x+6=0

a = -0.1; b = 0.8; c = +6;
Δ = b2-4ac
Δ = 0.82-4·(-0.1)·6
Δ = 3.04
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{3.04}}{2*-0.1}=\frac{-0.8-\sqrt{3.04}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{3.04}}{2*-0.1}=\frac{-0.8+\sqrt{3.04}}{-0.2}
$