(1/9)(6t-9)=t+(9/18)

Solution for (1/9)(6t-9)=t+(9/18) equation:

(1/9)(6t-9)=t+(9/18)

We move all terms to the left:

(1/9)(6t-9)-(t+(9/18))=0


Domain of the equation: 9)(6t-9)!=0

t∈R


We add all the numbers together, and all the variables

(+1/9)(6t-9)-(t+(+9/18))=0

We multiply parentheses ..

(+6t^2+1/9*-9)-(t+(+9/18))=0

We calculate fractions


6t^2/()+(-t)/()=0

We add all the numbers together, and all the variables

6t^2/()+(-1t)/()=0

We multiply all the terms by the denominator


6t^2+(-1t)=0

We get rid of parentheses

6t^2-1t=0

a = 6; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·6·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*6}=\frac{0}{12}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*6}=\frac{2}{12}
=1/6
$