-0.0009x^2+0.002x+45=0

Solution for -0.0009x^2+0.002x+45=0 equation:

-0.0009x^2+0.002x+45=0

a = -0.0009; b = 0.002; c = +45;
Δ = b2-4ac
Δ = 0.0022-4·(-0.0009)·45
Δ = 0.162004
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{-(0.002)-\sqrt{0.162004}}{2*-0.0009}=\frac{-0.002-\sqrt{0.162004}}{-0.0018}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.002)+\sqrt{0.162004}}{2*-0.0009}=\frac{-0.002+\sqrt{0.162004}}{-0.0018}
$