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14t-5t^2=0
a = -5; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-5)·0
Δ = 196
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{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-5}=\frac{-28}{-10} =2+4/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-5}=\frac{0}{-10} =0 $
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