5y^2-20y+17=0

Solution for 5y^2-20y+17=0 equation:

5y^2-20y+17=0

a = 5; b = -20; c = +17;
Δ = b2-4ac
Δ = -202-4·5·17
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{15}}{2*5}=\frac{20-2\sqrt{15}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{15}}{2*5}=\frac{20+2\sqrt{15}}{10}
$