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15t^2+19.5t=0
a = 15; b = 19.5; c = 0;
Δ = b2-4ac
Δ = 19.52-4·15·0
Δ = 380.25
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19.5)-\sqrt{380.25}}{2*15}=\frac{-19.5-\sqrt{380.25}}{30} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19.5)+\sqrt{380.25}}{2*15}=\frac{-19.5+\sqrt{380.25}}{30} $
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