-.3x^2+.1x=-.06

Solution for -.3x^2+.1x=-.06 equation:

-.3x^2+.1x=-.06

We move all terms to the left:

-.3x^2+.1x-(-.06)=0

We add all the numbers together, and all the variables

-.3x^2+.1x-(-0.06)=0

We add all the numbers together, and all the variables

-0.3x^2+.1x+0.06=0

a = -0.3; b = .1; c = +0.06;
Δ = b2-4ac
Δ = .12-4·(-0.3)·0.06
Δ = 0.082
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)-\sqrt{0.082}}{2*-0.3}=\frac{-0.1-\sqrt{0.082}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.1)+\sqrt{0.082}}{2*-0.3}=\frac{-0.1+\sqrt{0.082}}{-0.6}
$