x=3x^2+6x-4

Solution for x=3x^2+6x-4 equation:

x=3x^2+6x-4

We move all terms to the left:

x-(3x^2+6x-4)=0

We get rid of parentheses

-3x^2+x-6x+4=0

We add all the numbers together, and all the variables

-3x^2-5x+4=0

a = -3; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·(-3)·4
Δ = 73
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{-(-5)-\sqrt{73}}{2*-3}=\frac{5-\sqrt{73}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{73}}{2*-3}=\frac{5+\sqrt{73}}{-6}
$