10x-0.1x^2=560-0.2x^2

Solution for 10x-0.1x^2=560-0.2x^2 equation:

10x-0.1x^2=560-0.2x^2

We move all terms to the left:

10x-0.1x^2-(560-0.2x^2)=0

We get rid of parentheses

-0.1x^2+0.2x^2+10x-560=0

We add all the numbers together, and all the variables

0.1x^2+10x-560=0

a = 0.1; b = 10; c = -560;
Δ = b2-4ac
Δ = 102-4·0.1·(-560)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18}{2*0.1}=\frac{-28}{0.2}
=-140
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18}{2*0.1}=\frac{8}{0.2}
=40
$