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2y^2-9y-8=0
a = 2; b = -9; c = -8;
Δ = b2-4ac
Δ = -92-4·2·(-8)
Δ = 145
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{145}}{2*2}=\frac{9-\sqrt{145}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{145}}{2*2}=\frac{9+\sqrt{145}}{4} $
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