3x^2+8x+1=1

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

3x^2+8x+1=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

3x^2+8x=0

a = 3; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·3·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*3}=\frac{-16}{6}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*3}=\frac{0}{6}
=0
$