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12=32t-16t^2
We move all terms to the left:
12-(32t-16t^2)=0
We get rid of parentheses
16t^2-32t+12=0
a = 16; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·16·12
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*16}=\frac{16}{32} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*16}=\frac{48}{32} =1+1/2 $
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