16-2t=3/2t+

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

16-2t=3/2t+

We move all terms to the left:

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


Domain of the equation: 2t+)!=0

t∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-4t^2+32t-3=0

a = -4; b = 32; c = -3;
Δ = b2-4ac
Δ = 322-4·(-4)·(-3)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{61}}{2*-4}=\frac{-32-4\sqrt{61}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{61}}{2*-4}=\frac{-32+4\sqrt{61}}{-8}
$