b2/3+5=20-b

Solution for b2/3+5=20-b equation:

b2/3+5=20-b

We move all terms to the left:

b2/3+5-(20-b)=0

We add all the numbers together, and all the variables

b2/3-(-1b+20)+5=0

We get rid of parentheses

b2/3+1b-20+5=0

We multiply all the terms by the denominator


b2+1b*3-20*3+5*3=0

We add all the numbers together, and all the variables

b^2+1b*3-45=0

Wy multiply elements

b^2+3b-45=0

a = 1; b = 3; c = -45;
Δ = b2-4ac
Δ = 32-4·1·(-45)
Δ = 189
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:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{21}}{2*1}=\frac{-3-3\sqrt{21}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{21}}{2*1}=\frac{-3+3\sqrt{21}}{2}
$