(1/2t)+(2/3t)=35

Solution for (1/2t)+(2/3t)=35 equation:

(1/2t)+(2/3t)=35

We move all terms to the left:

(1/2t)+(2/3t)-(35)=0


Domain of the equation: 2t)!=0

t!=0/1

t!=0

t∈R



Domain of the equation: 3t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

(+1/2t)+(+2/3t)-35=0

We get rid of parentheses

1/2t+2/3t-35=0

We calculate fractions


3t/6t^2+4t/6t^2-35=0

We multiply all the terms by the denominator


3t+4t-35*6t^2=0

We add all the numbers together, and all the variables

7t-35*6t^2=0

Wy multiply elements

-210t^2+7t=0

a = -210; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-210)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-210}=\frac{-14}{-420}
=1/30
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-210}=\frac{0}{-420}
=0
$