3x^2-19x+3=0

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

3x^2-19x+3=0

a = 3; b = -19; c = +3;
Δ = b2-4ac
Δ = -192-4·3·3
Δ = 325
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:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5\sqrt{13}}{2*3}=\frac{19-5\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5\sqrt{13}}{2*3}=\frac{19+5\sqrt{13}}{6}
$