2(6x^2+19x-5)=0

Solution for 2(6x^2+19x-5)=0 equation:

2(6x^2+19x-5)=0

We multiply parentheses

12x^2+38x-10=0

a = 12; b = 38; c = -10;
Δ = b2-4ac
Δ = 382-4·12·(-10)
Δ = 1924
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:

$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{481}}{2*12}=\frac{-38-2\sqrt{481}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{481}}{2*12}=\frac{-38+2\sqrt{481}}{24}
$