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16y^2-5y-11=0
a = 16; b = -5; c = -11;
Δ = b2-4ac
Δ = -52-4·16·(-11)
Δ = 729
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{729}=27$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-27}{2*16}=\frac{-22}{32} =-11/16 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+27}{2*16}=\frac{32}{32} =1 $
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