16-2b=3/2b-9

Solution for 16-2b=3/2b-9 equation:

16-2b=3/2b-9

We move all terms to the left:

16-2b-(3/2b-9)=0


Domain of the equation: 2b-9)!=0

b∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-4b^2+50b-3=0

a = -4; b = 50; c = -3;
Δ = b2-4ac
Δ = 502-4·(-4)·(-3)
Δ = 2452
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2452}=\sqrt{4*613}=\sqrt{4}*\sqrt{613}=2\sqrt{613}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{613}}{2*-4}=\frac{-50-2\sqrt{613}}{-8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{613}}{2*-4}=\frac{-50+2\sqrt{613}}{-8}
$