(4/5)+(1/4)t=3

Solution for (4/5)+(1/4)t=3 equation:

(4/5)+(1/4)t=3

We move all terms to the left:

(4/5)+(1/4)t-(3)=0


Domain of the equation: 4)t!=0

t!=0/1

t!=0

t∈R


determiningTheFunctionDomain
(1/4)t-3+(4/5)=0

We add all the numbers together, and all the variables

(+1/4)t-3+(+4/5)=0

We multiply parentheses

t^2-3+(+4/5)=0

We get rid of parentheses

t^2-3+4/5=0

We multiply all the terms by the denominator


t^2*5+4-3*5=0

We add all the numbers together, and all the variables

t^2*5-11=0

Wy multiply elements

5t^2-11=0

a = 5; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·5·(-11)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{55}}{2*5}=\frac{0-2\sqrt{55}}{10}
=-\frac{2\sqrt{55}}{10}
=-\frac{\sqrt{55}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{55}}{2*5}=\frac{0+2\sqrt{55}}{10}
=\frac{2\sqrt{55}}{10}
=\frac{\sqrt{55}}{5}
$