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4y^2+18y-10=0
a = 4; b = 18; c = -10;
Δ = b2-4ac
Δ = 182-4·4·(-10)
Δ = 484
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{484}=22$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*4}=\frac{-40}{8} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*4}=\frac{4}{8} =1/2 $
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