4.9t^2+8t-40=0

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Solution for 4.9t^2+8t-40=0 equation:



4.9t^2+8t-40=0
a = 4.9; b = 8; c = -40;
Δ = b2-4ac
Δ = 82-4·4.9·(-40)
Δ = 848
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:
$\sqrt{\Delta}=\sqrt{848}=\sqrt{16*53}=\sqrt{16}*\sqrt{53}=4\sqrt{53}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{53}}{2*4.9}=\frac{-8-4\sqrt{53}}{9.8} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{53}}{2*4.9}=\frac{-8+4\sqrt{53}}{9.8} $

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