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3y^2+39y+120=0
a = 3; b = 39; c = +120;
Δ = b2-4ac
Δ = 392-4·3·120
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-9}{2*3}=\frac{-48}{6} =-8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+9}{2*3}=\frac{-30}{6} =-5 $
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