3x^2+8x+1=x

Solution for 3x^2+8x+1=x equation:

3x^2+8x+1=x

We move all terms to the left:

3x^2+8x+1-(x)=0

We add all the numbers together, and all the variables

3x^2+7x+1=0

a = 3; b = 7; c = +1;
Δ = b2-4ac
Δ = 72-4·3·1
Δ = 37
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{-(7)-\sqrt{37}}{2*3}=\frac{-7-\sqrt{37}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{37}}{2*3}=\frac{-7+\sqrt{37}}{6}
$