0,4x+0,2x^2=20

Solution for 0,4x+0,2x^2=20 equation:

0.4x+0.2x^2=20

We move all terms to the left:

0.4x+0.2x^2-(20)=0

a = 0.2; b = 0.4; c = -20;
Δ = b2-4ac
Δ = 0.42-4·0.2·(-20)
Δ = 16.16
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{-(0.4)-\sqrt{16.16}}{2*0.2}=\frac{-0.4-\sqrt{16.16}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{16.16}}{2*0.2}=\frac{-0.4+\sqrt{16.16}}{0.4}
$