2/3b+5=20+b

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Solution for 2/3b+5=20+b equation:



2/3b+5=20+b
We move all terms to the left:
2/3b+5-(20+b)=0
Domain of the equation: 3b!=0
b!=0/3
b!=0
b∈R
We add all the numbers together, and all the variables
2/3b-(b+20)+5=0
We get rid of parentheses
2/3b-b-20+5=0
We multiply all the terms by the denominator
-b*3b-20*3b+5*3b+2=0
Wy multiply elements
-3b^2-60b+15b+2=0
We add all the numbers together, and all the variables
-3b^2-45b+2=0
a = -3; b = -45; c = +2;
Δ = b2-4ac
Δ = -452-4·(-3)·2
Δ = 2049
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{2049}}{2*-3}=\frac{45-\sqrt{2049}}{-6} $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{2049}}{2*-3}=\frac{45+\sqrt{2049}}{-6} $

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