-0.3x^2+0.5x+6=0

Solution for -0.3x^2+0.5x+6=0 equation:

-0.3x^2+0.5x+6=0

a = -0.3; b = 0.5; c = +6;
Δ = b2-4ac
Δ = 0.52-4·(-0.3)·6
Δ = 7.45
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{-(0.5)-\sqrt{7.45}}{2*-0.3}=\frac{-0.5-\sqrt{7.45}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{7.45}}{2*-0.3}=\frac{-0.5+\sqrt{7.45}}{-0.6}
$