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2t^2-19.8t-14.88=0
a = 2; b = -19.8; c = -14.88;
Δ = b2-4ac
Δ = -19.82-4·2·(-14.88)
Δ = 511.08
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{-(-19.8)-\sqrt{511.08}}{2*2}=\frac{19.8-\sqrt{511.08}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.8)+\sqrt{511.08}}{2*2}=\frac{19.8+\sqrt{511.08}}{4} $
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