3x^2-16x+17=0

Solution for 3x^2-16x+17=0 equation:

3x^2-16x+17=0

a = 3; b = -16; c = +17;
Δ = b2-4ac
Δ = -162-4·3·17
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{-(-16)-2\sqrt{13}}{2*3}=\frac{16-2\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{13}}{2*3}=\frac{16+2\sqrt{13}}{6}
$