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49(t)=4.9t^2
We move all terms to the left:
49(t)-(4.9t^2)=0
We get rid of parentheses
-4.9t^2+49t=0
a = -4.9; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·(-4.9)·0
Δ = 2401
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*-4.9}=\frac{-98}{-9.8} =+10 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*-4.9}=\frac{0}{-9.8} =0 $
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