2x(20x+1)=11

Solution for 2x(20x+1)=11 equation:

2x(20x+1)=11

We move all terms to the left:

2x(20x+1)-(11)=0

We multiply parentheses

40x^2+2x-11=0

a = 40; b = 2; c = -11;
Δ = b2-4ac
Δ = 22-4·40·(-11)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-42}{2*40}=\frac{-44}{80}
=-11/20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+42}{2*40}=\frac{40}{80}
=1/2
$