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4.9t^2+21t-5=0
a = 4.9; b = 21; c = -5;
Δ = b2-4ac
Δ = 212-4·4.9·(-5)
Δ = 539
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{539}=\sqrt{49*11}=\sqrt{49}*\sqrt{11}=7\sqrt{11}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-7\sqrt{11}}{2*4.9}=\frac{-21-7\sqrt{11}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+7\sqrt{11}}{2*4.9}=\frac{-21+7\sqrt{11}}{9.8} $
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