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1.5t^2+4.9t+2=0

a = 1.5; b = 4.9; c = +2;

Δ = b^{2}-4ac

Δ = 4.9^{2}-4·1.5·2

Δ = 12.01

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{-(4.9)-\sqrt{12.01}}{2*1.5}=\frac{-4.9-\sqrt{12.01}}{3} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.9)+\sqrt{12.01}}{2*1.5}=\frac{-4.9+\sqrt{12.01}}{3} $

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