0,0438=0,1x+0,25x^2

Solution for 0,0438=0,1x+0,25x^2 equation:

0.0438=0.1x+0.25x^2

We move all terms to the left:

0.0438-(0.1x+0.25x^2)=0

We get rid of parentheses

-0.25x^2-0.1x+0.0438=0

a = -0.25; b = -0.1; c = +0.0438;
Δ = b2-4ac
Δ = -0.12-4·(-0.25)·0.0438
Δ = 0.0538
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{-(-0.1)-\sqrt{0.0538}}{2*-0.25}=\frac{0.1-\sqrt{0.0538}}{-0.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1)+\sqrt{0.0538}}{2*-0.25}=\frac{0.1+\sqrt{0.0538}}{-0.5}
$