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6y^2+31y+40=0
a = 6; b = 31; c = +40;
Δ = b2-4ac
Δ = 312-4·6·40
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*6}=\frac{-32}{12} =-2+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*6}=\frac{-30}{12} =-2+1/2 $
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