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24y^2+18y-42=0
a = 24; b = 18; c = -42;
Δ = b2-4ac
Δ = 182-4·24·(-42)
Δ = 4356
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{4356}=66$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-66}{2*24}=\frac{-84}{48} =-1+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+66}{2*24}=\frac{48}{48} =1 $
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