-0.005x^2+x-3.2=0

Solution for -0.005x^2+x-3.2=0 equation:

-0.005x^2+x-3.2=0

a = -0.005; b = 1; c = -3.2;
Δ = b2-4ac
Δ = 12-4·(-0.005)·(-3.2)
Δ = 0.936
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{0.936}}{2*-0.005}=\frac{-1-\sqrt{0.936}}{-0.01}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.936}}{2*-0.005}=\frac{-1+\sqrt{0.936}}{-0.01}
$