1/5t-4=-2/3t+2

Solution for 1/5t-4=-2/3t+2 equation:

1/5t-4=-2/3t+2

We move all terms to the left:

1/5t-4-(-2/3t+2)=0


Domain of the equation: 5t!=0

t!=0/5

t!=0

t∈R



Domain of the equation: 3t+2)!=0

t∈R


We get rid of parentheses

1/5t+2/3t-2-4=0

We calculate fractions


3t/15t^2+10t/15t^2-2-4=0

We add all the numbers together, and all the variables

3t/15t^2+10t/15t^2-6=0

We multiply all the terms by the denominator


3t+10t-6*15t^2=0

We add all the numbers together, and all the variables

13t-6*15t^2=0

Wy multiply elements

-90t^2+13t=0

a = -90; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-90)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-90}=\frac{-26}{-180}
=13/90
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-90}=\frac{0}{-180}
=0
$