-.005x^2+1.2x+3=0

Solution for -.005x^2+1.2x+3=0 equation:

-.005x^2+1.2x+3=0

We add all the numbers together, and all the variables

-0.005x^2+1.2x+3=0

a = -0.005; b = 1.2; c = +3;
Δ = b2-4ac
Δ = 1.22-4·(-0.005)·3
Δ = 1.5
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{1.5}}{2*-0.005}=\frac{-1.2-\sqrt{1.5}}{-0.01}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{1.5}}{2*-0.005}=\frac{-1.2+\sqrt{1.5}}{-0.01}
$