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

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



-(t/2)+7=-(1/3)t+4
We move all terms to the left:
-(t/2)+7-(-(1/3)t+4)=0
Domain of the equation: 3)t+4)!=0
t!=0/1
t!=0
t∈R
We add all the numbers together, and all the variables
-(+t/2)-(-(+1/3)t+4)+7=0
We get rid of parentheses
-t/2-(-(+1/3)t+4)+7=0
We calculate fractions
(-3t^2)/6t+()/6t+7=0
We multiply all the terms by the denominator
(-3t^2)+7*6t+()=0
We add all the numbers together, and all the variables
(-3t^2)+7*6t=0
Wy multiply elements
(-3t^2)+42t=0
We get rid of parentheses
-3t^2+42t=0
a = -3; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·(-3)·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*-3}=\frac{-84}{-6} =+14 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*-3}=\frac{0}{-6} =0 $

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