.15x=2+.12x-0.002x^2

Solution for .15x=2+.12x-0.002x^2 equation:

.15x=2+.12x-0.002x^2

We move all terms to the left:

.15x-(2+.12x-0.002x^2)=0

We get rid of parentheses

0.002x^2-.12x+.15x-2=0

We add all the numbers together, and all the variables

0.002x^2+0.03x-2=0

a = 0.002; b = 0.03; c = -2;
Δ = b2-4ac
Δ = 0.032-4·0.002·(-2)
Δ = 0.0169
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{-(0.03)-\sqrt{0.0169}}{2*0.002}=\frac{-0.03-\sqrt{0.0169}}{0.004}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.03)+\sqrt{0.0169}}{2*0.002}=\frac{-0.03+\sqrt{0.0169}}{0.004}
$