18+1/3t=16+7/10t

Solution for 18+1/3t=16+7/10t equation:

18+1/3t=16+7/10t

We move all terms to the left:

18+1/3t-(16+7/10t)=0


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R



Domain of the equation: 10t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

1/3t-(7/10t+16)+18=0

We get rid of parentheses

1/3t-7/10t-16+18=0

We calculate fractions


10t/30t^2+(-21t)/30t^2-16+18=0

We add all the numbers together, and all the variables

10t/30t^2+(-21t)/30t^2+2=0

We multiply all the terms by the denominator


10t+(-21t)+2*30t^2=0

Wy multiply elements

60t^2+10t+(-21t)=0

We get rid of parentheses

60t^2+10t-21t=0

We add all the numbers together, and all the variables

60t^2-11t=0

a = 60; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·60·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*60}=\frac{0}{120}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*60}=\frac{22}{120}
=11/60
$