16-2t=3/4t+9

Solution for 16-2t=3/4t+9 equation:

16-2t=3/4t+9

We move all terms to the left:

16-2t-(3/4t+9)=0


Domain of the equation: 4t+9)!=0

t∈R


We get rid of parentheses

-2t-3/4t-9+16=0

We multiply all the terms by the denominator


-2t*4t-9*4t+16*4t-3=0

Wy multiply elements

-8t^2-36t+64t-3=0

We add all the numbers together, and all the variables

-8t^2+28t-3=0

a = -8; b = 28; c = -3;
Δ = b2-4ac
Δ = 282-4·(-8)·(-3)
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{43}}{2*-8}=\frac{-28-4\sqrt{43}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{43}}{2*-8}=\frac{-28+4\sqrt{43}}{-16}
$