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-3t^2+24t+60=0
a = -3; b = 24; c = +60;
Δ = b2-4ac
Δ = 242-4·(-3)·60
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-36}{2*-3}=\frac{-60}{-6} =+10 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+36}{2*-3}=\frac{12}{-6} =-2 $
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