2(3x^2+4)=19

Solution for 2(3x^2+4)=19 equation:

2(3x^2+4)=19

We move all terms to the left:

2(3x^2+4)-(19)=0

We multiply parentheses

6x^2+8-19=0

We add all the numbers together, and all the variables

6x^2-11=0

a = 6; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·6·(-11)
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{66}}{2*6}=\frac{0-2\sqrt{66}}{12}
=-\frac{2\sqrt{66}}{12}
=-\frac{\sqrt{66}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{66}}{2*6}=\frac{0+2\sqrt{66}}{12}
=\frac{2\sqrt{66}}{12}
=\frac{\sqrt{66}}{6}
$