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13t+6=5t^2
We move all terms to the left:
13t+6-(5t^2)=0
determiningTheFunctionDomain -5t^2+13t+6=0
a = -5; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·(-5)·6
Δ = 289
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{289}=17$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-5}=\frac{-30}{-10} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-5}=\frac{4}{-10} =-2/5 $
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