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

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

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

We move all terms to the left:

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

We multiply parentheses

6x-6x-(6x^2+1)+2+12=0

We get rid of parentheses

-6x^2+6x-6x-1+2+12=0

We add all the numbers together, and all the variables

-6x^2+13=0

a = -6; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-6)·13
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{78}}{2*-6}=\frac{0-2\sqrt{78}}{-12}
=-\frac{2\sqrt{78}}{-12}
=-\frac{\sqrt{78}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{78}}{2*-6}=\frac{0+2\sqrt{78}}{-12}
=\frac{2\sqrt{78}}{-12}
=\frac{\sqrt{78}}{-6}
$