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10y^2-9y-9=0
a = 10; b = -9; c = -9;
Δ = b2-4ac
Δ = -92-4·10·(-9)
Δ = 441
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{441}=21$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-21}{2*10}=\frac{-12}{20} =-3/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+21}{2*10}=\frac{30}{20} =1+1/2 $
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