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4.9t^2-20t-40=0
a = 4.9; b = -20; c = -40;
Δ = b2-4ac
Δ = -202-4·4.9·(-40)
Δ = 1184
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1184}=\sqrt{16*74}=\sqrt{16}*\sqrt{74}=4\sqrt{74}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{74}}{2*4.9}=\frac{20-4\sqrt{74}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{74}}{2*4.9}=\frac{20+4\sqrt{74}}{9.8} $
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