-0.15x^2+x+0.5=0

Solution for -0.15x^2+x+0.5=0 equation:

-0.15x^2+x+0.5=0

a = -0.15; b = 1; c = +0.5;
Δ = b2-4ac
Δ = 12-4·(-0.15)·0.5
Δ = 1.3
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{-(1)-\sqrt{1.3}}{2*-0.15}=\frac{-1-\sqrt{1.3}}{-0.3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1.3}}{2*-0.15}=\frac{-1+\sqrt{1.3}}{-0.3}
$