0,1x^2+0,9x=20

Solution for 0,1x^2+0,9x=20 equation:

0.1x^2+0.9x=20

We move all terms to the left:

0.1x^2+0.9x-(20)=0

a = 0.1; b = 0.9; c = -20;
Δ = b2-4ac
Δ = 0.92-4·0.1·(-20)
Δ = 8.81
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{-(0.9)-\sqrt{8.81}}{2*0.1}=\frac{-0.9-\sqrt{8.81}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.9)+\sqrt{8.81}}{2*0.1}=\frac{-0.9+\sqrt{8.81}}{0.2}
$