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

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

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

We move all terms to the left:

80+4x+0.1x^2-(10x)=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{4}{0.2}
=20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*0.1}=\frac{8}{0.2}
=40
$