3y^2+24=-18y

Solution for 3y^2+24=-18y equation:

3y^2+24=-18y

We move all terms to the left:

3y^2+24-(-18y)=0

We get rid of parentheses

3y^2+18y+24=0

a = 3; b = 18; c = +24;
Δ = b2-4ac
Δ = 182-4·3·24
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*3}=\frac{-24}{6}
=-4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*3}=\frac{-12}{6}
=-2
$