y^2-18y+62=0

Solution for y^2-18y+62=0 equation:

y^2-18y+62=0

a = 1; b = -18; c = +62;
Δ = b2-4ac
Δ = -182-4·1·62
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{19}}{2*1}=\frac{18-2\sqrt{19}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{19}}{2*1}=\frac{18+2\sqrt{19}}{2}
$