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8y^2-41y-5=0
a = 8; b = -41; c = -5;
Δ = b2-4ac
Δ = -412-4·8·(-5)
Δ = 1841
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{-(-41)-\sqrt{1841}}{2*8}=\frac{41-\sqrt{1841}}{16} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{1841}}{2*8}=\frac{41+\sqrt{1841}}{16} $
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