41x+20=-20x^2

Solution for 41x+20=-20x^2 equation:

41x+20=-20x^2

We move all terms to the left:

41x+20-(-20x^2)=0

We get rid of parentheses

20x^2+41x+20=0

a = 20; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·20·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9}{2*20}=\frac{-50}{40}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9}{2*20}=\frac{-32}{40}
=-4/5
$