(14/5)+(1/6)t=3

Solution for (14/5)+(1/6)t=3 equation:

(14/5)+(1/6)t=3

We move all terms to the left:

(14/5)+(1/6)t-(3)=0


Domain of the equation: 6)t!=0

t!=0/1

t!=0

t∈R


determiningTheFunctionDomain
(1/6)t-3+(14/5)=0

We add all the numbers together, and all the variables

(+1/6)t-3+(+14/5)=0

We multiply parentheses

t^2-3+(+14/5)=0

We get rid of parentheses

t^2-3+14/5=0

We multiply all the terms by the denominator


t^2*5+14-3*5=0

We add all the numbers together, and all the variables

t^2*5-1=0

Wy multiply elements

5t^2-1=0

a = 5; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·5·(-1)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*5}=\frac{0-2\sqrt{5}}{10}
=-\frac{2\sqrt{5}}{10}
=-\frac{\sqrt{5}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*5}=\frac{0+2\sqrt{5}}{10}
=\frac{2\sqrt{5}}{10}
=\frac{\sqrt{5}}{5}
$