2x^2-18x+34=0

Solution for 2x^2-18x+34=0 equation:

2x^2-18x+34=0

a = 2; b = -18; c = +34;
Δ = b2-4ac
Δ = -182-4·2·34
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{13}}{2*2}=\frac{18-2\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{13}}{2*2}=\frac{18+2\sqrt{13}}{4}
$