y^2+14y+20=0

Solution for y^2+14y+20=0 equation:

y^2+14y+20=0

a = 1; b = 14; c = +20;
Δ = b2-4ac
Δ = 142-4·1·20
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{29}}{2*1}=\frac{-14-2\sqrt{29}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{29}}{2*1}=\frac{-14+2\sqrt{29}}{2}
$