2/3t+8/9t=98

Solution for 2/3t+8/9t=98 equation:

2/3t+8/9t=98

We move all terms to the left:

2/3t+8/9t-(98)=0


Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R



Domain of the equation: 9t!=0

t!=0/9

t!=0

t∈R


We calculate fractions


18t/27t^2+24t/27t^2-98=0

We multiply all the terms by the denominator


18t+24t-98*27t^2=0

We add all the numbers together, and all the variables

42t-98*27t^2=0

Wy multiply elements

-2646t^2+42t=0

a = -2646; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·(-2646)·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*-2646}=\frac{-84}{-5292}
=1/63
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*-2646}=\frac{0}{-5292}
=0
$