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3y^2+28y-20=0
a = 3; b = 28; c = -20;
Δ = b2-4ac
Δ = 282-4·3·(-20)
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-32}{2*3}=\frac{-60}{6} =-10 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+32}{2*3}=\frac{4}{6} =2/3 $
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