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6.5t^2+13t=0
a = 6.5; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·6.5·0
Δ = 169
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{169}=13$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*6.5}=\frac{-26}{13} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*6.5}=\frac{0}{13} =0 $
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