2x^2-18x+17=0

Solution for 2x^2-18x+17=0 equation:

2x^2-18x+17=0

a = 2; b = -18; c = +17;
Δ = b2-4ac
Δ = -182-4·2·17
Δ = 188
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:

$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{47}}{2*2}=\frac{18-2\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{47}}{2*2}=\frac{18+2\sqrt{47}}{4}
$