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4.9t^2-11t-4=0
a = 4.9; b = -11; c = -4;
Δ = b2-4ac
Δ = -112-4·4.9·(-4)
Δ = 199.4
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{-(-11)-\sqrt{199.4}}{2*4.9}=\frac{11-\sqrt{199.4}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{199.4}}{2*4.9}=\frac{11+\sqrt{199.4}}{9.8} $
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