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

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
$