4y^2-18y+17=0

Solution for 4y^2-18y+17=0 equation:

4y^2-18y+17=0

a = 4; b = -18; c = +17;
Δ = b2-4ac
Δ = -182-4·4·17
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{13}}{2*4}=\frac{18-2\sqrt{13}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{13}}{2*4}=\frac{18+2\sqrt{13}}{8}
$