2x(3x+4)=19

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

2x(3x+4)=19

We move all terms to the left:

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

We multiply parentheses

6x^2+8x-19=0

a = 6; b = 8; c = -19;
Δ = b2-4ac
Δ = 82-4·6·(-19)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{130}}{2*6}=\frac{-8-2\sqrt{130}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{130}}{2*6}=\frac{-8+2\sqrt{130}}{12}
$