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4.9t^2-12t=80
We move all terms to the left:
4.9t^2-12t-(80)=0
a = 4.9; b = -12; c = -80;
Δ = b2-4ac
Δ = -122-4·4.9·(-80)
Δ = 1712
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1712}=\sqrt{16*107}=\sqrt{16}*\sqrt{107}=4\sqrt{107}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{107}}{2*4.9}=\frac{12-4\sqrt{107}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{107}}{2*4.9}=\frac{12+4\sqrt{107}}{9.8} $
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