3x^2+1=9x^2+6x+1

Solution for 3x^2+1=9x^2+6x+1 equation:

3x^2+1=9x^2+6x+1

We move all terms to the left:

3x^2+1-(9x^2+6x+1)=0

We get rid of parentheses

3x^2-9x^2-6x-1+1=0

We add all the numbers together, and all the variables

-6x^2-6x=0

a = -6; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-6)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-6}=\frac{0}{-12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-6}=\frac{12}{-12}
=-1
$