3x^2+4x-19=0

Solution for 3x^2+4x-19=0 equation:

3x^2+4x-19=0

a = 3; b = 4; c = -19;
Δ = b2-4ac
Δ = 42-4·3·(-19)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{61}}{2*3}=\frac{-4-2\sqrt{61}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{61}}{2*3}=\frac{-4+2\sqrt{61}}{6}
$