y^2+22y+29=0

Solution for y^2+22y+29=0 equation:

y^2+22y+29=0

a = 1; b = 22; c = +29;
Δ = b2-4ac
Δ = 222-4·1·29
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-4\sqrt{23}}{2*1}=\frac{-22-4\sqrt{23}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+4\sqrt{23}}{2*1}=\frac{-22+4\sqrt{23}}{2}
$