(9/5)+(1/2)t=7

Solution for (9/5)+(1/2)t=7 equation:

(9/5)+(1/2)t=7

We move all terms to the left:

(9/5)+(1/2)t-(7)=0


Domain of the equation: 2)t!=0

t!=0/1

t!=0

t∈R


determiningTheFunctionDomain
(1/2)t-7+(9/5)=0

We add all the numbers together, and all the variables

(+1/2)t-7+(+9/5)=0

We multiply parentheses

t^2-7+(+9/5)=0

We get rid of parentheses

t^2-7+9/5=0

We multiply all the terms by the denominator


t^2*5+9-7*5=0

We add all the numbers together, and all the variables

t^2*5-26=0

Wy multiply elements

5t^2-26=0

a = 5; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·5·(-26)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{130}}{2*5}=\frac{0-2\sqrt{130}}{10}
=-\frac{2\sqrt{130}}{10}
=-\frac{\sqrt{130}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{130}}{2*5}=\frac{0+2\sqrt{130}}{10}
=\frac{2\sqrt{130}}{10}
=\frac{\sqrt{130}}{5}
$