0=3x^2+4x+1=

Solution for 0=3x^2+4x+1= equation:

0=3x^2+4x+1=

We move all terms to the left:

0-(3x^2+4x+1)=0

We add all the numbers together, and all the variables

-(3x^2+4x+1)=0

We get rid of parentheses

-3x^2-4x-1=0

a = -3; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·(-3)·(-1)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*-3}=\frac{2}{-6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*-3}=\frac{6}{-6}
=-1
$