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12y^2+13y-103=0
a = 12; b = 13; c = -103;
Δ = b2-4ac
Δ = 132-4·12·(-103)
Δ = 5113
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{-(13)-\sqrt{5113}}{2*12}=\frac{-13-\sqrt{5113}}{24} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{5113}}{2*12}=\frac{-13+\sqrt{5113}}{24} $
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