20+1.2x-0.002x^2=0

Solution for 20+1.2x-0.002x^2=0 equation:

20+1.2x-0.002x^2=0

a = -0.002; b = 1.2; c = +20;
Δ = b2-4ac
Δ = 1.22-4·(-0.002)·20
Δ = 1.6
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{-(1.2)-\sqrt{1.6}}{2*-0.002}=\frac{-1.2-\sqrt{1.6}}{-0.004}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{1.6}}{2*-0.002}=\frac{-1.2+\sqrt{1.6}}{-0.004}
$