(7/3)t-2=19+5t

Solution for (7/3)t-2=19+5t equation:

(7/3)t-2=19+5t

We move all terms to the left:

(7/3)t-2-(19+5t)=0


Domain of the equation: 3)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+7/3)t-(5t+19)-2=0

We multiply parentheses

7t^2-(5t+19)-2=0

We get rid of parentheses

7t^2-5t-19-2=0

We add all the numbers together, and all the variables

7t^2-5t-21=0

a = 7; b = -5; c = -21;
Δ = b2-4ac
Δ = -52-4·7·(-21)
Δ = 613
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{613}}{2*7}=\frac{5-\sqrt{613}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{613}}{2*7}=\frac{5+\sqrt{613}}{14}
$