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-16t^2+5t+2.5=0
a = -16; b = 5; c = +2.5;
Δ = b2-4ac
Δ = 52-4·(-16)·2.5
Δ = 185
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{-(5)-\sqrt{185}}{2*-16}=\frac{-5-\sqrt{185}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{185}}{2*-16}=\frac{-5+\sqrt{185}}{-32} $
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