16-2b=3/2b-9

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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} $

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