3x+8=x+20x2

Solution for 3x+8=x+20x2 equation:

3x+8=x+20x^2

We move all terms to the left:

3x+8-(x+20x^2)=0

We get rid of parentheses

-20x^2-x+3x+8=0

We add all the numbers together, and all the variables

-20x^2+2x+8=0

a = -20; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-20)·8
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{161}}{2*-20}=\frac{-2-2\sqrt{161}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{161}}{2*-20}=\frac{-2+2\sqrt{161}}{-40}
$