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4.9t^2+5t-30=0
a = 4.9; b = 5; c = -30;
Δ = b2-4ac
Δ = 52-4·4.9·(-30)
Δ = 613
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{613}}{2*4.9}=\frac{-5-\sqrt{613}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{613}}{2*4.9}=\frac{-5+\sqrt{613}}{9.8} $
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