x=-0.02x^2+0.8x+10

Solution for x=-0.02x^2+0.8x+10 equation:

x=-0.02x^2+0.8x+10

We move all terms to the left:

x-(-0.02x^2+0.8x+10)=0

We get rid of parentheses

0.02x^2-0.8x+x-10=0

We add all the numbers together, and all the variables

0.02x^2+0.2x-10=0

a = 0.02; b = 0.2; c = -10;
Δ = b2-4ac
Δ = 0.22-4·0.02·(-10)
Δ = 0.84
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{-(0.2)-\sqrt{0.84}}{2*0.02}=\frac{-0.2-\sqrt{0.84}}{0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{0.84}}{2*0.02}=\frac{-0.2+\sqrt{0.84}}{0.04}
$