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-16t^2+48t=-16t
We move all terms to the left:
-16t^2+48t-(-16t)=0
We get rid of parentheses
-16t^2+48t+16t=0
We add all the numbers together, and all the variables
-16t^2+64t=0
a = -16; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·(-16)·0
Δ = 4096
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{4096}=64$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*-16}=\frac{-128}{-32} =+4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*-16}=\frac{0}{-32} =0 $
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