T-5(1/6x-8)=40-x-1/6x

Solution for T-5(1/6x-8)=40-x-1/6x equation:

-5(1/6T-8)=40-T-1/6T

We move all terms to the left:

-5(1/6T-8)-(40-T-1/6T)=0


Domain of the equation: 6T-8)!=0

T∈R



Domain of the equation: 6T)!=0

T!=0/1

T!=0

T∈R


We add all the numbers together, and all the variables

-5(1/6T-8)-(-1T-1/6T+40)=0

We multiply parentheses

-5T-(-1T-1/6T+40)+40=0

We get rid of parentheses

-5T+1T+1/6T-40+40=0

We multiply all the terms by the denominator


-5T*6T+1T*6T-40*6T+40*6T+1=0

Wy multiply elements

-30T^2+6T^2-240T+240T+1=0

We add all the numbers together, and all the variables

-24T^2+1=0

a = -24; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-24)·1
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-24}=\frac{0-4\sqrt{6}}{-48}
=-\frac{4\sqrt{6}}{-48}
=-\frac{\sqrt{6}}{-12}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-24}=\frac{0+4\sqrt{6}}{-48}
=\frac{4\sqrt{6}}{-48}
=\frac{\sqrt{6}}{-12}
$