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12y^2-3y-8=0
a = 12; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·12·(-8)
Δ = 393
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{393}}{2*12}=\frac{3-\sqrt{393}}{24} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{393}}{2*12}=\frac{3+\sqrt{393}}{24} $
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