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3y^2+22y+24=0
a = 3; b = 22; c = +24;
Δ = b2-4ac
Δ = 222-4·3·24
Δ = 196
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{196}=14$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*3}=\frac{-36}{6} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*3}=\frac{-8}{6} =-1+1/3 $
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