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3=5t^2+3
We move all terms to the left:
3-(5t^2+3)=0
We get rid of parentheses
-5t^2-3+3=0
We add all the numbers together, and all the variables
-5t^2=0
a = -5; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-5)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$t=\frac{-b}{2a}=\frac{0}{-10}=0$
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