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

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

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

We move all terms to the left:

(7/3t)-2-(4+7t)=0


Domain of the equation: 3t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+7/3t)-(7t+4)-2=0

We get rid of parentheses

7/3t-7t-4-2=0

We multiply all the terms by the denominator


-7t*3t-4*3t-2*3t+7=0

Wy multiply elements

-21t^2-12t-6t+7=0

We add all the numbers together, and all the variables

-21t^2-18t+7=0

a = -21; b = -18; c = +7;
Δ = b2-4ac
Δ = -182-4·(-21)·7
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{57}}{2*-21}=\frac{18-4\sqrt{57}}{-42}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{57}}{2*-21}=\frac{18+4\sqrt{57}}{-42}
$