12x^2+4x-19=0

Solution for 12x^2+4x-19=0 equation:

12x^2+4x-19=0

a = 12; b = 4; c = -19;
Δ = b2-4ac
Δ = 42-4·12·(-19)
Δ = 928
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:

$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{58}}{2*12}=\frac{-4-4\sqrt{58}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{58}}{2*12}=\frac{-4+4\sqrt{58}}{24}
$