3x^2-22x+19=0

Solution for 3x^2-22x+19=0 equation:

3x^2-22x+19=0

a = 3; b = -22; c = +19;
Δ = b2-4ac
Δ = -222-4·3·19
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-16}{2*3}=\frac{6}{6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+16}{2*3}=\frac{38}{6}
=6+1/3
$