y^2+20y+46=0

Solution for y^2+20y+46=0 equation:

y^2+20y+46=0

a = 1; b = 20; c = +46;
Δ = b2-4ac
Δ = 202-4·1·46
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-6\sqrt{6}}{2*1}=\frac{-20-6\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+6\sqrt{6}}{2*1}=\frac{-20+6\sqrt{6}}{2}
$