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80t-16t^2=96
We move all terms to the left:
80t-16t^2-(96)=0
a = -16; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·(-16)·(-96)
Δ = 256
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-16}{2*-16}=\frac{-96}{-32} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+16}{2*-16}=\frac{-64}{-32} =+2 $
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