x=3x^2-18x+29

Solution for x=3x^2-18x+29 equation:

x=3x^2-18x+29

We move all terms to the left:

x-(3x^2-18x+29)=0

We get rid of parentheses

-3x^2+x+18x-29=0

We add all the numbers together, and all the variables

-3x^2+19x-29=0

a = -3; b = 19; c = -29;
Δ = b2-4ac
Δ = 192-4·(-3)·(-29)
Δ = 13
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{-(19)-\sqrt{13}}{2*-3}=\frac{-19-\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{13}}{2*-3}=\frac{-19+\sqrt{13}}{-6}
$