3x^2-22x+34=0

Solution for 3x^2-22x+34=0 equation:

3x^2-22x+34=0

a = 3; b = -22; c = +34;
Δ = b2-4ac
Δ = -222-4·3·34
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{19}}{2*3}=\frac{22-2\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{19}}{2*3}=\frac{22+2\sqrt{19}}{6}
$