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2.5t^2-5t-2.25=0
a = 2.5; b = -5; c = -2.25;
Δ = b2-4ac
Δ = -52-4·2.5·(-2.25)
Δ = 47.5
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{47.5}}{2*2.5}=\frac{5-\sqrt{47.5}}{5} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{47.5}}{2*2.5}=\frac{5+\sqrt{47.5}}{5} $
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