3y^2=-22y-35

Solution for 3y^2=-22y-35 equation:

3y^2=-22y-35

We move all terms to the left:

3y^2-(-22y-35)=0

We get rid of parentheses

3y^2+22y+35=0

a = 3; b = 22; c = +35;
Δ = b2-4ac
Δ = 222-4·3·35
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-8}{2*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*3}=\frac{-14}{6}
=-2+1/3
$