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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

6x^2-19x+8=0

a = 6; b = -19; c = +8;
Δ = b2-4ac
Δ = -192-4·6·8
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*6}=\frac{6}{12}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*6}=\frac{32}{12}
=2+2/3
$