20=0.8x^2+10x

Solution for 20=0.8x^2+10x equation:

20=0.8x^2+10x

We move all terms to the left:

20-(0.8x^2+10x)=0

We get rid of parentheses

-0.8x^2-10x+20=0

a = -0.8; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·(-0.8)·20
Δ = 164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*-0.8}=\frac{10-2\sqrt{41}}{-1.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*-0.8}=\frac{10+2\sqrt{41}}{-1.6}
$