y^2+22y-5=0

Solution for y^2+22y-5=0 equation:

y^2+22y-5=0

a = 1; b = 22; c = -5;
Δ = b2-4ac
Δ = 222-4·1·(-5)
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-6\sqrt{14}}{2*1}=\frac{-22-6\sqrt{14}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+6\sqrt{14}}{2*1}=\frac{-22+6\sqrt{14}}{2}
$