16t^2+48t+40=4

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Solution for 16t^2+48t+40=4 equation: 



16t^2+48t+40=4

We move all terms to the left:

16t^2+48t+40-(4)=0

We add all the numbers together, and all the variables

16t^2+48t+36=0

a = 16; b = 48; c = +36;
Δ = b2-4ac
Δ = 482-4·16·36
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-48}{32}=-1+1/2$

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