(7t^2-4t+3)=t^2+1-1

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Solution for (7t^2-4t+3)=t^2+1-1 equation: 



(7t^2-4t+3)=t^2+1-1

We move all terms to the left:

(7t^2-4t+3)-(t^2+1-1)=0

We get rid of parentheses

7t^2-t^2-4t+3-1+1=0

We add all the numbers together, and all the variables

6t^2-4t+3=0

a = 6; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·6·3
Δ = -56
Delta is less than zero, so there is no solution for the equation

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