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12c^2+23c-24=0
a = 12; b = 23; c = -24;
Δ = b2-4ac
Δ = 232-4·12·(-24)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1681}=41$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-41}{2*12}=\frac{-64}{24} =-2+2/3 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+41}{2*12}=\frac{18}{24} =3/4 $
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