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3=3+95t-16t^2
We move all terms to the left:
3-(3+95t-16t^2)=0
We get rid of parentheses
16t^2-95t-3+3=0
We add all the numbers together, and all the variables
16t^2-95t=0
a = 16; b = -95; c = 0;
Δ = b2-4ac
Δ = -952-4·16·0
Δ = 9025
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{9025}=95$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-95}{2*16}=\frac{0}{32} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+95}{2*16}=\frac{190}{32} =5+15/16 $
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