16-2t=3/2t+5

Solution for 16-2t=3/2t+5 equation:

16-2t=3/2t+5

We move all terms to the left:

16-2t-(3/2t+5)=0


Domain of the equation: 2t+5)!=0

t∈R


We get rid of parentheses

-2t-3/2t-5+16=0

We multiply all the terms by the denominator


-2t*2t-5*2t+16*2t-3=0

Wy multiply elements

-4t^2-10t+32t-3=0

We add all the numbers together, and all the variables

-4t^2+22t-3=0

a = -4; b = 22; c = -3;
Δ = b2-4ac
Δ = 222-4·(-4)·(-3)
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{109}}{2*-4}=\frac{-22-2\sqrt{109}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{109}}{2*-4}=\frac{-22+2\sqrt{109}}{-8}
$