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-3t^2+42t-39=0
a = -3; b = 42; c = -39;
Δ = b2-4ac
Δ = 422-4·(-3)·(-39)
Δ = 1296
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-36}{2*-3}=\frac{-78}{-6} =+13 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+36}{2*-3}=\frac{-6}{-6} =1 $
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