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12X^2+13X-19=0
a = 12; b = 13; c = -19;
Δ = b2-4ac
Δ = 132-4·12·(-19)
Δ = 1081
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{1081}}{2*12}=\frac{-13-\sqrt{1081}}{24} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{1081}}{2*12}=\frac{-13+\sqrt{1081}}{24} $
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