2x^2-16x+17=0

Solution for 2x^2-16x+17=0 equation:

2x^2-16x+17=0

a = 2; b = -16; c = +17;
Δ = b2-4ac
Δ = -162-4·2·17
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{30}}{2*2}=\frac{16-2\sqrt{30}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{30}}{2*2}=\frac{16+2\sqrt{30}}{4}
$