0=t^2+4t+12

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Solution for 0=t^2+4t+12 equation: 



0=t^2+4t+12

We move all terms to the left:

0-(t^2+4t+12)=0

We add all the numbers together, and all the variables

-(t^2+4t+12)=0

We get rid of parentheses

-t^2-4t-12=0

We add all the numbers together, and all the variables

-1t^2-4t-12=0

a = -1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·(-1)·(-12)
Δ = -32
Delta is less than zero, so there is no solution for the equation

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