20=0.01308x^2+1.1439x

Solution for 20=0.01308x^2+1.1439x equation:

20=0.01308x^2+1.1439x

We move all terms to the left:

20-(0.01308x^2+1.1439x)=0

We get rid of parentheses

-0.01308x^2-1.1439x+20=0

a = -0.01308; b = -1.1439; c = +20;
Δ = b2-4ac
Δ = -1.14392-4·(-0.01308)·20
Δ = 2.35490721
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{-(-1.1439)-\sqrt{2.35490721}}{2*-0.01308}=\frac{1.1439-\sqrt{2.35490721}}{-0.02616}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.1439)+\sqrt{2.35490721}}{2*-0.01308}=\frac{1.1439+\sqrt{2.35490721}}{-0.02616}
$