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4.9t^2+3.5t-2.4=0
a = 4.9; b = 3.5; c = -2.4;
Δ = b2-4ac
Δ = 3.52-4·4.9·(-2.4)
Δ = 59.29
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{-(3.5)-\sqrt{59.29}}{2*4.9}=\frac{-3.5-\sqrt{59.29}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.5)+\sqrt{59.29}}{2*4.9}=\frac{-3.5+\sqrt{59.29}}{9.8} $
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