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20y^2-3y-8=0
a = 20; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·20·(-8)
Δ = 649
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{-(-3)-\sqrt{649}}{2*20}=\frac{3-\sqrt{649}}{40} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{649}}{2*20}=\frac{3+\sqrt{649}}{40} $
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