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40t+4.9t^2=50
We move all terms to the left:
40t+4.9t^2-(50)=0
a = 4.9; b = 40; c = -50;
Δ = b2-4ac
Δ = 402-4·4.9·(-50)
Δ = 2580
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{645}}{2*4.9}=\frac{-40-2\sqrt{645}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{645}}{2*4.9}=\frac{-40+2\sqrt{645}}{9.8} $
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