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3y^2+13y+12=0
a = 3; b = 13; c = +12;
Δ = b2-4ac
Δ = 132-4·3·12
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*3}=\frac{-18}{6} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*3}=\frac{-8}{6} =-1+1/3 $
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