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

Solution for (-4/5)t+(2/5)=(2/3) equation:

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

We move all terms to the left:

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


Domain of the equation: 5)t!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

-4t^2+2=0

a = -4; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-4)·2
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-4}=\frac{0-4\sqrt{2}}{-8}
=-\frac{4\sqrt{2}}{-8}
=-\frac{\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-4}=\frac{0+4\sqrt{2}}{-8}
=\frac{4\sqrt{2}}{-8}
=\frac{\sqrt{2}}{-2}
$