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4.9t^2+11t-0.4=0
a = 4.9; b = 11; c = -0.4;
Δ = b2-4ac
Δ = 112-4·4.9·(-0.4)
Δ = 128.84
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{128.84}}{2*4.9}=\frac{-11-\sqrt{128.84}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{128.84}}{2*4.9}=\frac{-11+\sqrt{128.84}}{9.8} $
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