16-2t=3/9t+9

Solution for 16-2t=3/9t+9 equation:

16-2t=3/9t+9

We move all terms to the left:

16-2t-(3/9t+9)=0


Domain of the equation: 9t+9)!=0

t∈R


We get rid of parentheses

-2t-3/9t-9+16=0

We multiply all the terms by the denominator


-2t*9t-9*9t+16*9t-3=0

Wy multiply elements

-18t^2-81t+144t-3=0

We add all the numbers together, and all the variables

-18t^2+63t-3=0

a = -18; b = 63; c = -3;
Δ = b2-4ac
Δ = 632-4·(-18)·(-3)
Δ = 3753
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3753}=\sqrt{9*417}=\sqrt{9}*\sqrt{417}=3\sqrt{417}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-3\sqrt{417}}{2*-18}=\frac{-63-3\sqrt{417}}{-36}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+3\sqrt{417}}{2*-18}=\frac{-63+3\sqrt{417}}{-36}
$