2(3x^2-7x+1)=4

Solution for 2(3x^2-7x+1)=4 equation:

2(3x^2-7x+1)=4

We move all terms to the left:

2(3x^2-7x+1)-(4)=0

We multiply parentheses

6x^2-14x+2-4=0

We add all the numbers together, and all the variables

6x^2-14x-2=0

a = 6; b = -14; c = -2;
Δ = b2-4ac
Δ = -142-4·6·(-2)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{61}}{2*6}=\frac{14-2\sqrt{61}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{61}}{2*6}=\frac{14+2\sqrt{61}}{12}
$