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12t^2-5t-4=0
a = 12; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·12·(-4)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{217}}{2*12}=\frac{5-\sqrt{217}}{24} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{217}}{2*12}=\frac{5+\sqrt{217}}{24} $
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