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-4.9t^2+4t+152.4=0
a = -4.9; b = 4; c = +152.4;
Δ = b2-4ac
Δ = 42-4·(-4.9)·152.4
Δ = 3003.04
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{-(4)-\sqrt{3003.04}}{2*-4.9}=\frac{-4-\sqrt{3003.04}}{-9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{3003.04}}{2*-4.9}=\frac{-4+\sqrt{3003.04}}{-9.8} $
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