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(3+2t)-(8+1)=2/t
We move all terms to the left:
(3+2t)-(8+1)-(2/t)=0
Domain of the equation: t)!=0We add all the numbers together, and all the variables
t!=0/1
t!=0
t∈R
(2t+3)-(+2/t)-9=0
We get rid of parentheses
2t-2/t+3-9=0
We multiply all the terms by the denominator
2t*t+3*t-9*t-2=0
We add all the numbers together, and all the variables
-6t+2t*t-2=0
Wy multiply elements
2t^2-6t-2=0
a = 2; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·2·(-2)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*2}=\frac{6-2\sqrt{13}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*2}=\frac{6+2\sqrt{13}}{4} $
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