2x(10X+1)=5

Solution for 2x(10X+1)=5 equation:

2x(10x+1)=5

We move all terms to the left:

2x(10x+1)-(5)=0

We multiply parentheses

20x^2+2x-5=0

a = 20; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·20·(-5)
Δ = 404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{101}}{2*20}=\frac{-2-2\sqrt{101}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{101}}{2*20}=\frac{-2+2\sqrt{101}}{40}
$