17y^2-62y+49=0

Solution for 17y^2-62y+49=0 equation:

17y^2-62y+49=0

a = 17; b = -62; c = +49;
Δ = b2-4ac
Δ = -622-4·17·49
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-16\sqrt{2}}{2*17}=\frac{62-16\sqrt{2}}{34}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+16\sqrt{2}}{2*17}=\frac{62+16\sqrt{2}}{34}
$