.0013(.16-x)=x^2+.16x

Solution for .0013(.16-x)=x^2+.16x equation:

.0013(.16-x)=x^2+.16x

We move all terms to the left:

.0013(.16-x)-(x^2+.16x)=0

We add all the numbers together, and all the variables

.0013(-1x+0.16)-(x^2+.16x)=0

We multiply parentheses

-0.0013x-(x^2+.16x)+0.000208=0

We get rid of parentheses

-x^2-0.0013x-.16x+0.000208=0

We add all the numbers together, and all the variables

-1x^2-0.1613x+0.000208=0

a = -1; b = -0.1613; c = +0.000208;
Δ = b2-4ac
Δ = -0.16132-4·(-1)·0.000208
Δ = 0.02684969
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{-(-0.1613)-\sqrt{0.02684969}}{2*-1}=\frac{0.1613-\sqrt{0.02684969}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1613)+\sqrt{0.02684969}}{2*-1}=\frac{0.1613+\sqrt{0.02684969}}{-2}
$