(3x^2-3x+4)+(3x^2+3x-6)=1

Solution for (3x^2-3x+4)+(3x^2+3x-6)=1 equation:

(3x^2-3x+4)+(3x^2+3x-6)=1

We move all terms to the left:

(3x^2-3x+4)+(3x^2+3x-6)-(1)=0

We get rid of parentheses

3x^2+3x^2-3x+3x+4-6-1=0

We add all the numbers together, and all the variables

6x^2-3=0

a = 6; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·6·(-3)
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*6}=\frac{0-6\sqrt{2}}{12}
=-\frac{6\sqrt{2}}{12}
=-\frac{\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*6}=\frac{0+6\sqrt{2}}{12}
=\frac{6\sqrt{2}}{12}
=\frac{\sqrt{2}}{2}
$