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4.9t^2+17t+10=0
a = 4.9; b = 17; c = +10;
Δ = b2-4ac
Δ = 172-4·4.9·10
Δ = 93
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{-(17)-\sqrt{93}}{2*4.9}=\frac{-17-\sqrt{93}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{93}}{2*4.9}=\frac{-17+\sqrt{93}}{9.8} $
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