x(3x-4)+1=0

Solution for x(3x-4)+1=0 equation:

x(3x-4)+1=0

We multiply 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
$