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2y^2+35y-18=0
a = 2; b = 35; c = -18;
Δ = b2-4ac
Δ = 352-4·2·(-18)
Δ = 1369
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{1369}=37$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-37}{2*2}=\frac{-72}{4} =-18 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+37}{2*2}=\frac{2}{4} =1/2 $
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