K(3)=t2+5

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Solution for K(3)=t2+5 equation: 



(3)=K2+5

We move all terms to the left:

(3)-(K2+5)=0

We add all the numbers together, and all the variables

-(+K^2+5)+3=0

We get rid of parentheses

-K^2-5+3=0

We add all the numbers together, and all the variables

-1K^2-2=0

a = -1; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·(-1)·(-2)
Δ = -8
Delta is less than zero, so there is no solution for the equation

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