16+2t=3/2t+9

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

16+2t=3/2t+9

We move all terms to the left:

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


Domain of the equation: 2t+9)!=0

t∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

4t^2-18t+32t-3=0

We add all the numbers together, and all the variables

4t^2+14t-3=0

a = 4; b = 14; c = -3;
Δ = b2-4ac
Δ = 142-4·4·(-3)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{61}}{2*4}=\frac{-14-2\sqrt{61}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{61}}{2*4}=\frac{-14+2\sqrt{61}}{8}
$