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4.9t^2=50
We move all terms to the left:
4.9t^2-(50)=0
a = 4.9; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·4.9·(-50)
Δ = 980
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{980}=\sqrt{4*245}=\sqrt{4}*\sqrt{245}=2\sqrt{245}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{245}}{2*4.9}=\frac{0-2\sqrt{245}}{9.8} =-\frac{2\sqrt{245}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{245}}{2*4.9}=\frac{0+2\sqrt{245}}{9.8} =\frac{2\sqrt{245}}{9.8} $
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