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5/2t+-t=3+3/2t
We move all terms to the left:
5/2t+-t-(3+3/2t)=0
Domain of the equation: 2t!=0
t!=0/2
t!=0
t∈R
Domain of the equation: 2t)!=0We add all the numbers together, and all the variables
t!=0/1
t!=0
t∈R
5/2t-t-(3/2t+3)+=0
We add all the numbers together, and all the variables
-1t+5/2t-(3/2t+3)=0
We get rid of parentheses
-1t+5/2t-3/2t-3=0
We multiply all the terms by the denominator
-1t*2t-3*2t+5-3=0
We add all the numbers together, and all the variables
-1t*2t-3*2t+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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