3y^2+36y+63=0

Solution for 3y^2+36y+63=0 equation:

3y^2+36y+63=0

a = 3; b = 36; c = +63;
Δ = b2-4ac
Δ = 362-4·3·63
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{15}}{2*3}=\frac{-36-6\sqrt{15}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{15}}{2*3}=\frac{-36+6\sqrt{15}}{6}
$