(10^-3)x=0.012

Solution for (10^-3)x=0.012 equation:

(10^-3)x=0.012

We move all terms to the left:

(10^-3)x-(0.012)=0

We add all the numbers together, and all the variables

(10^-3)x-0.012=0

We multiply parentheses

10x^2-3x-0.012=0

a = 10; b = -3; c = -0.012;
Δ = b2-4ac
Δ = -32-4·10·(-0.012)
Δ = 9.48
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{-(-3)-\sqrt{9.48}}{2*10}=\frac{3-\sqrt{9.48}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{9.48}}{2*10}=\frac{3+\sqrt{9.48}}{20}
$