x^2-19x+34=0.

Solution for x^2-19x+34=0. equation:

x^2-19x+34=0.

a = 1; b = -19; c = +34;
Δ = b2-4ac
Δ = -192-4·1·34
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-15}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+15}{2*1}=\frac{34}{2}
=17
$