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4.9t^2+28t-170=0
a = 4.9; b = 28; c = -170;
Δ = b2-4ac
Δ = 282-4·4.9·(-170)
Δ = 4116
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4116}=\sqrt{196*21}=\sqrt{196}*\sqrt{21}=14\sqrt{21}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-14\sqrt{21}}{2*4.9}=\frac{-28-14\sqrt{21}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+14\sqrt{21}}{2*4.9}=\frac{-28+14\sqrt{21}}{9.8} $
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