10x=80+4x+0.1x^2

Solution for 10x=80+4x+0.1x^2 equation:

10x=80+4x+0.1x^2

We move all terms to the left:

10x-(80+4x+0.1x^2)=0

We get rid of parentheses

-0.1x^2-4x+10x-80=0

We add all the numbers together, and all the variables

-0.1x^2+6x-80=0

a = -0.1; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·(-0.1)·(-80)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-0.1}=\frac{-8}{-0.2}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-0.1}=\frac{-4}{-0.2}
=+20
$