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3y^2+2y-56=0
a = 3; b = 2; c = -56;
Δ = b2-4ac
Δ = 22-4·3·(-56)
Δ = 676
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{676}=26$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-26}{2*3}=\frac{-28}{6} =-4+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+26}{2*3}=\frac{24}{6} =4 $
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